Hc Verma I for Class 12 Science Physics Chapter 15 - Wave Motion And Wave On A String

Answer:

No, in wave motion there is no actual transfer of matter but transfer of energy between the points where as when wind blows air particles moves with it.

Answer:

It is a non-mechanical wave because this type of wave does not require a material medium to travel.

Answer:

Equation of the wave is
$y=c_1 \sin \left(c_{2}x+c_{3}t\right)$
When the variable of the equation is (c₂x + c₃t), then the wave must be moving in the negative x-axis with time t.

Answer:

Equation of the wave is given by
$y=A\sin \left(\omega t-kx\right)$
where
          A is the amplitude
           ω is the angular frequency
           k is the wave number
Velocity of wave, $v=\frac{\omega }{k}$
Velocity of particle, $v_p=\frac{dy}{dt}=A\omega \cos \left(\omega t-kx\right)$
Max velocity of particle, $v_{p_{\max }}=A\omega$
As given
$A<\frac{\lambda }{2\mathrm{\pi }}$
$$\begin{gathered} v_{p_{\max }}=\frac{\lambda \omega }{2\mathrm{\pi }} \\ {v}_{p_{\max }}<\frac{\omega }{k} \left[\because \frac{2\mathrm{\pi }}{\lambda }=k\right] \end{gathered}$$

Answer:

When two wave pulses identical in shape but inverted with respect to each other meet at any instant, they form a destructive interference. The complete energy of the system at that instant is stored in the form of potential energy within it. After passing each other, both the pulses regain their original shape.

Answer:

$$\begin{gathered} \frac{y_{\max }}{v_{\max }}=\frac{v_{\max }}{a_{\max }} \\ y=A\sin \left(\omega t-kx\right) \\ y= A \\ v=\frac{dy}{dt}=A\omega \cos \left(\omega t-kx\right) \\ {v}_{\max }=A\omega \\ a=\frac{dv}{dt}=-A{\omega }^2\sin \left(\omega t-kx\right) \\ {a}_{\max }={\omega }^{2}A \end{gathered}$$

To prove,

$$\begin{gathered} \frac{y_{\max }}{v_{\max }}=\frac{v_{\max }}{a_{\max }} \\ \mathrm{LHS} \\ \frac{y_{\max }}{v_{\max }}=\frac{A}{A\omega }=\frac{1}{\omega } \\ \mathrm{RHS} \\ \frac{v_{\max }}{a_{\max }}=\frac{A\omega }{{\omega }^{2}A}=\frac{1}{\omega } \end{gathered}$$

No, componendo and dividendo is not applicable. We cannot add quantities of different dimensions.

Answer:

Equation of the wave: y = A sin(kx − ωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.

The argument of the sine is a phase, so the smallest positive phase constant should be          
$$\begin{gathered} \sin \left(7.5\mathrm{\pi }\right)=\sin \left(3\times 2\mathrm{\pi }+1.5\mathrm{\pi }\right) \\ =\sin \left(1.5\mathrm{\pi }\right) \end{gathered}$$
Therefore, the smallest positive phase constant is 1.5π.

Answer:

Yes, at the centre. The centre position is a node. If the string vibrates in its first overtone, then there will be two positions, i.e., two nodes, one at x = 0 and the other at x = L.

Answer:

(c) $\lambda /2$

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the positions of maxima and minima is equal to $\lambda /2$.

Answer:

(c) $\lambda /2$

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant, then the particle closest to it having zero displacement is at a distance is equal to $\lambda /2$.

Answer:

(a) $x=A \sin \left(ky-\omega t\right)$

Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.

Answer:

(b) amplitude A/2, frequency $\omega /\pi$

$y=A{\sin }^2\left(kx-\omega t\right)$  

$$\begin{gathered} \left[{\cos }^2\theta =1-2{\sin }^2\theta \\ {\sin }^2\theta =\frac{1-{\cos }^2\theta }{2}\right] \end{gathered}$$

$$\begin{gathered} y=A\left[\frac{1-{\cos }^2\left(kx-\omega t\right)}{2}\right] \\ y=\frac{A}{2}\left[1-{\cos }^2\left(kx-\omega t\right)\right] \end{gathered}$$
Thus, we have:
Amplitude = $\frac{A}{2}$
Frequency = $2\left(\frac{\omega }{2\pi }\right)=\frac{\omega }{\pi }$

Answer:

(d) Sound waves

There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound requires medium to travel, hence it is a mechanical wave.

Answer:

(a) $\nu$

The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.

Answer:

(a) 1/2

Wave speed is given by
$\nu =\sqrt{\frac{T}{\mathit{\mu }}}$
where
             ​T is the tension in the string
             v is the speed of the wave
             μ is the mass per unit length of the string
$\mu =\frac{M}{L}=\rho \frac{V}{L}=\rho \frac{\left(AL\right)}{L}$
where
           M is the mass of the string, which can be written as ρV
             L is the length of the string
$$\begin{gathered} =\rho \left(\pi r^2\right)=\rho \left(\pi \frac{D^2}{4}\right) \\ \therefore \nu =\sqrt{\frac{T}{\rho \pi {\displaystyle \frac{D^2}{4}}}}=\frac{2}{D}\sqrt{\frac{T}{\rho \pi }} \end{gathered}$$
where D is the diameter of the string.
Thus, v ∝ $\frac{1}{D}$
Since, r_A = 2r_B
$$\begin{gathered} v_{\mathrm{A}}\propto \frac{1}{2r_{\mathrm{A}}}\propto \frac{1}{2\times 2r_{\mathrm{B}}} \left(1\right) \\ {v}_{{}_{\mathrm{B}}}\propto \frac{1}{2r_{{}_{\mathrm{B}}}} \left(2\right) \end{gathered}$$
From Equations (1) and (2) we get
$\frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{1}{2}$

Answer:

(d) $1/\sqrt{2}$


$$\begin{gathered} T_{\mathrm{AB}}=T \\ {T}_{\mathrm{CD}}=2T \end{gathered}$$
where
          ​T_AB is the tension in the string AB
          ​T_CD is the tension in the string CD
The eelation between tension and the wave speed is given by
$$\begin{gathered} v=\sqrt{\frac{T}{\mathit{\mu }}} \\ v\propto \sqrt{T} \end{gathered}$$
where
          v is the wave speed of the transverse wave
           μ is the mass per unit length of the string
$\frac{v_1}{v_2}=\sqrt{\frac{T}{2T}}=\frac{1}{\sqrt{2}}$

Answer:

(d) meaningless

Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.

Answer:

(c) $\lambda \text{'}<\lambda$

As $v=\sqrt{\frac{f}{\mathit{\mu }}}$
A wave pulse travels faster in a thinner string.
The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.

Answer:

(b) $\sqrt{2}a$

We know that the resultant of the amplitude is given by
$R_{\mathrm{net}}=\sqrt{A_1^2+A_2^2+2A_1{A}_2 \cos \varphi }$
For the particular case, we can write
$$\begin{gathered} =\sqrt{a^2+a^2+2a^2\cos \frac{\mathrm{\pi }}{2}} \\ =\sqrt{2}a \end{gathered}$$

Answer:

(d) the information is insufficient to find the relation between t₁ and t₂.

$v=\sqrt{\frac{\eta }{\rho }}$
But because the length of wires A and B is not known, the relation between A and B cannot be determined.

Answer:

(b) the velocity but not for the kinetic energy

The principle of superposition is valid only for vector quantities. Velocity is a vector quantity, but kinetic energy is a scalar quantity.

Answer:

(d) The pulses will pass through each other without any change in their shapes.

The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.

Answer:

(b) 2A₂

We know resultant amplitude is given by
$A_{\mathrm{net}}=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$
For maximum resultant amplitude
$A_{\max }=A_1+A_2$
For minimum resultant amplitude
$A_{\min }=A_1-A_2$

So, the difference between A_max and A_min is
$A_{\max }-A_{\min }=A_1+A_2-A_1+A_2=2A_2$

Answer:

(d) between 0 and 2A

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle (φ). So, the resultant amplitude lies between the maximum resultant amplitude (A_max) and the minimum resultant amplitude (A_min).
A_max = A + A = 2A
A_min = A − A = 0

Answer:

(a) A

We know the resultant amplitude is given by
$$\begin{gathered} R_{\mathrm{net}}=\sqrt{A^2+A^2+2A^2 \cos 120º} (\varphi =120°) \\ =\sqrt{2A^2-A^2} \left[\because \cos 120º=\frac{-1}{2}\right] \\ =A \end{gathered}$$

Answer:

(a) inverse of its length

The relation between wave speed and the length of the string is given by
$v=\frac{1}{2l}\sqrt{\frac{F}{\mathit{\mu }}}$
where
           l is the length of the string
           F is the tension
           μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
$v\propto \frac{1}{l}$
       

Answer:

Given,
Speed of the wave pulse passing on a string in the negative x-direction = 40 cms⁻¹
As the speed of the wave is constant, the location of the maximum after 5 s will be
s = v × t
   = 40 × 5
   = 200 cm (along the negative x-axis)
Therefore, the required maximum will be located after x = −2 m.

Answer:

Given,
Equation of the wave travelling on a string stretched along the X-axis:
$y=\mathrm{A}e{ }^{\left(\frac{x}{a}+\frac{t}{T}\right){{}^{-}}^2}$
(a) The dimensions of A (amplitude), T (time period) and $a=\frac{\lambda }{2\pi }$, which will have the dimensions of the wavelength, are as follows:
$$\begin{gathered} \left[\mathrm{A}\right] = \left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right] \\ \left[\mathrm{T}\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right] \\ \left[a\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right] \end{gathered}$$

(b) Wave speed, $\nu =\frac{\lambda }{T}=\frac{a}{T} \left[\lambda =a\right]$

(c) If $y=f \left(t+\frac{x}{\nu }\right)$, then the wave travels in the negative direction; and if $y=f \left(t-\frac{x}{\nu }\right)$, then the wave travels in the positive direction.
Thus, we have:
$$\begin{gathered} y=Ae{ }^{{{\left[\left(\frac{x}{a}\right)+\left(\frac{t}{T}\right)\right]}^{-}}^2} \\ =A{e}^{-\frac{1}{T}{\left[t+\left(\frac{xT}{a}\right)\right]}^2} \\ =A{e}^{-\frac{1}{T}\left[t+\frac{x}{V}\right]} \\ = A{e}^{-f\left[t+\frac{x}{V}\right] } \end{gathered}$$
Hence, the wave is travelling is the negative direction.

(d) Wave speed, $v=\frac{a}{t}$
Maximum pulse at t = T  = $\left(\frac{a}{T}\right)\times T=a \left(\mathrm{Along} \mathrm{the} \mathrm{negative} x-\mathrm{axis}\right)$
Maximum pulse at t = 2T = $\left(\frac{a}{T}\times 2T\right)=2a \left(\mathrm{Along} \mathrm{the} \mathrm{negative} x-\mathrm{axis}\right)$
Therefore, the wave is travelling in the negative x-direction.

Answer:

Given,
Wave pulse at t = 0

Wave speed = 10 cms⁻¹
Using the formula $s=v\times t$, we get:

$$\begin{gathered} \mathrm{At}: \\ t=1 \mathrm{s}, s_1=\nu \times t=10\times 1=10 \mathrm{cm} \\ t = 2 \mathrm{s}, s_2=\nu \times t=10\times 2=20 \mathrm{cm} \\ t = 3 \mathrm{s}, s_3=\nu \times t=10\times 3=30 \mathrm{cm} \end{gathered}$$

Answer:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.

Answer:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.

Answer:

(b) vibrate with a frequency of 208 Hz

According to the relation of the fundamental frequency of a string
$\nu =\frac{1}{2l}\sqrt{\frac{F}{\mu }}$
where
          l is the length of the string
           F is the tension
           μ is the linear mass density

We know that ν₁ = 416 Hz, l₁ = l and l₂ = 2l.
$$\begin{gathered} v_1\propto \frac{1}{l_1} \\ {v}_1{l}_1=v_2{l}_2 \\ \left(416\right)l=v_2\left(2l\right) \\ {v}_2=208 \mathrm{Hz} \end{gathered}$$

Answer:

(d) 16 kg

According to the relation of the fundamental frequency of a string
$\nu =\frac{1}{2l}\sqrt{\frac{F}{\mu }}$
where l is the length of the string
           F is the tension
           μ is the linear mass density of the string
We know that ν₁ = 416 Hz, l₁ = l and l₂ = 2l.
Also, m₁ = 4 kg and m₂ = ? 
${\nu }_1=\frac{1}{2l_1}\sqrt{\frac{m_{1}g}{\mu }} \left(1\right)$
${\nu }_2=\frac{1}{2l_2}\sqrt{\frac{m_{2}g}{\mu }} \left(2\right)$
So, in order to maintain the same fundamental mode
${\nu }_1={\nu }_2$
squaring both sides of equations (1) and (2) and then equating
$$\begin{gathered} \frac{1}{4l^2}\frac{4g}{\mu } = \frac{1}{16l^2}\frac{m_{2}g}{\mu } \\ \Rightarrow m_2=16 \mathrm{kg} \end{gathered}$$

Answer:

(c) may move on the X-axis
(d) may move on the Y-axis

A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.

Answer:

(d) in the X–Y plane

In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the X–Y plane.

Answer:

(d) be polarised

A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.

Answer:

(b) may be longitudinal
(d) may be transverse

Particles in a solid are very close to each other; thus, both longitudinal and transverse waves can travel through it.

Answer:

(a) must be longitudinal

Because particles in a gas are far apart, only longitudinal wave can travel through it.

Answer:

(b) A and B move in opposite directions.
(d) The displacements at A and B have equal magnitudes.


A and B have a phase difference of π. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.
$\overrightarrow{y_A}=A\sin \left(\omega t\right)$
$\overrightarrow{y_B}=B\sin \left(\omega t+\mathrm{\pi }\right)$

Answer:

(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm

$y=\left(0·001 \mathrm{mm}\right) \sin \left[\left(50 {\mathrm{s}}^{-1}\right)t+\left(2·0 {\mathrm{m}}^{-1}\right)x\right]$
Equating the above equation with the general equation, we get:
$$\begin{gathered} y=A\sin \left(\omega t-kx\right) \\ \omega =\frac{2\mathrm{\pi }}{T}=2\pi v \\ k=\frac{2\mathrm{\pi }}{\mathrm{\lambda }} \end{gathered}$$
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.
$$\begin{gathered} A=0.001 \mathrm{mm} \\ \mathrm{Now}, \\ 50=2\pi \nu \\ \Rightarrow \mathrm{\nu }=\frac{25}{\mathrm{\pi }} \mathrm{Hz} \end{gathered}$$

Answer:

(a) must be an integral multiple of $\lambda /4$

A standing wave is produced on a string clamped at one end and free at the other. Its fundamental frequency is given by
$$\begin{gathered} \nu =\left(n+\frac{1}{2}\right)\frac{v}{2L} \\ \Rightarrow v=\nu \lambda \\ \Rightarrow \nu =\left(n+\frac{1}{2}\right)\frac{\nu \lambda }{2L} \\ \Rightarrow L=\left(\frac{2n+1}{4}\right)\lambda \\ \Rightarrow L=\frac{\lambda }{4},\frac{3\lambda }{4},... \end{gathered}$$

Answer:

(b) The energy of any small part of a string remains constant in a standing wave.

A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.

Answer:

(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase

All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.

Thus, particles in alternate antinodes vibrate in the same phase.

Answer:

Given,
Pulse travelling on a string, $y=\left[\frac{{\left(a\right)}^3}{{\left(x-\nu t\right)}^2+a^2}\right]$
$$\begin{gathered} a=5 \mathrm{mm}=0.5 \mathrm{cm} \\ \mathrm{Wave} \mathrm{speed}, \nu = 20 \mathrm{cm}/\mathrm{s} \end{gathered}$$
So, at $t=0 \mathrm{s}, y=\frac{a^3}{\left(x^2+a^2\right)}$.
Similarly, at t = 1 s,
$$\begin{gathered} y=\frac{a^3}{{\left(x-\nu \right)}^2+a^2} \\ \mathrm{And}, \\ \mathrm{At} t=2 \mathrm{s}, \\ y=\frac{a^3}{{\left(x-2\nu \right)}^2+a^2} \end{gathered}$$

To sketch the shape of the string, we have to plot a graph between y and x at different values of t.

Answer:

Given,
Equation of the wave travelling in the positive x-direction at x = 0:
$f\left(t\right)=A\sin \left(\frac{t}{T}\right)$
Here,
Wave speed = v
Wavelength, λ = vT
T = Time period
Therefore, the general equation of the wave can be represented by
$y=A\sin \left[\left(\frac{t}{T}\right)-\left(\frac{x}{\nu T}\right)\right]$

Answer:

The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
Dimensions of A and a are governed by the dimensional homogeneity of the equation g(x) = A sin(x/a).
Now,
$$\begin{gathered} \left(a\right) \left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]=\left[\mathrm{A}\right] \\ \Rightarrow \left[A\right]=\left[\mathrm{L}\right] \\ \mathrm{And}, \\ \left[a\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right] \\ \Rightarrow \left[a\right]=\left[\mathrm{L}\right] \\ \left(b\right) \mathrm{Wave} \mathrm{speed}=\nu \\ \therefore \mathrm{Time} \mathrm{period}, T=\frac{a}{\nu } \\ \mathrm{Here}, \\ a=\mathrm{Wave} \mathrm{length}=\lambda \\ \mathrm{The} \mathrm{general} \mathrm{equation} \mathrm{of} \mathrm{wave} \mathrm{is} \mathrm{represented} \mathrm{by} \\ y=A\sin \left\{\frac{x}{a}-\frac{t}{{\displaystyle \frac{{\displaystyle a}}{v}}}\right\} \\ =A\sin \left\{\frac{x-\nu t}{a}\right\} \end{gathered}$$

Answer:

Given,
Wave velocity = $\nu$
Shape of the string at $t=t_0$ = $g\left(x, t_0\right)=A \sin \left(x/a\right)$    ...(i)
For a wave travelling in the positive x-direction, the general equation is given by
$y=A \sin \left(\frac{x}{a}-\frac{t}{T}\right)$
Putting t = − t and comparing with equation (i), we get:
$$\begin{gathered} g\left(x,0\right)=A\sin \left\{\left(\frac{x}{a}\right)+\left(\frac{t_0}{T}\right)\right\} \\ \Rightarrow g\left(x,t\right)=A\sin \left[\left\{\left(\frac{x}{a}\right)+\frac{t_0}{T}\right\}-\left(\frac{t}{T}\right)\right] \\ \mathrm{Now}, \\ T=\frac{a}{\nu } \\ \mathrm{Here}, \\ a=\mathrm{Wave} \mathrm{length} \\ \nu =\mathrm{Velocity} \mathrm{of} \mathrm{the} \mathrm{wave} \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ y=A\sin \left[\left(\frac{x}{a}\right)+\frac{t_0}{\left({\displaystyle \frac{a}{\nu }}\right)}-\frac{t}{\left({\displaystyle \frac{a}{\nu }}\right)}\right] \end{gathered}$$

$\Rightarrow y=A\sin \frac{x+\nu \left(t_0-t\right)}{a}$

Answer:

Given,
Equation of the wave, $y=\left(0.10 \mathrm{mm}\right) \sin \left(31.4 {\mathrm{m}}^{-1}\right)x+\left(314 {\mathrm{s}}^{-1}\right) t$
The general equation is $y=A\sin \left\{\left(\frac{2\pi x}{\lambda }\right)+\omega t\right\}$.
From the above equation, we can conclude:

(a) The wave is travelling in the negative x-direction.

(b) $\frac{2\pi }{\lambda }=31.4 {\mathrm{m}}^{-1}$
$$\begin{gathered} \Rightarrow \lambda =\frac{2\pi }{31.4}=0.2 \mathrm{m}= 20 \mathrm{cm} \\ \mathrm{And}, \\ \omega =314 {\mathrm{s}}^{-1} \\ \Rightarrow 2\pi f=314 \\ \Rightarrow f=\frac{314}{2\pi } \\ =\frac{314}{2\times 3.14} \\ =50 {\mathrm{s}}^{-1}=50 \mathrm{Hz} \end{gathered}$$

Wave speed:
$$\begin{gathered} \nu =\lambda f=20\times 50 \\ = 1000 \mathrm{cm}/\mathrm{s} \end{gathered}$$

(c) Maximum displacement, A = 0.10 mm
$$\begin{gathered} \mathrm{Maximum} \mathrm{velocity}=a\omega =0.1\times {10}^{-1}\times 314 \\ =3.14 \mathrm{cm}/\mathrm{s} \end{gathered}$$

Answer:

A wave travels along the positive x-direction.
Wave amplitude (A) = 0.20 cm
Wavelength (λ) = 20 cm
Wave speed (v) = 20 m/s

(a) General wave equation along the x-axis:
$$\begin{gathered} y=A\sin \left(kx-\omega t\right) \\ \therefore k=\frac{2\pi }{\lambda }=\frac{2\pi }{2}=\pi {\mathrm{cm}}^{-1} \\ T=\frac{\lambda }{\nu }=\frac{2}{2000} \\ =\frac{1}{1000}={10}^{-3} \mathrm{s} \\ \omega =\frac{2\pi }{T}=2\pi \times {10}^3 {\mathrm{s}}^{-1} \end{gathered}$$
Wave equation:
$y=\left(0.2 \mathrm{cm}\right) \sin \left[\left(\pi {\mathrm{cm}}^{-1}\right) x-\left(2\pi \times {10}^{-3} {\mathrm{s}}^{-1}\right)\right]$

(b) As per the question
For the wave equation ,we need to find the displacement and velocity at x = 2 cm and t = 0.
$$\begin{gathered} y=\left(0.2\right) \mathrm{cm} \sin 2\pi =0 \\ \therefore \nu =A\omega \mathrm{cos\pi }x \\ =0.2\times 2000\pi \times \cos 2\pi \\ =400\pi \\ =400\pi \mathrm{cm}/\mathrm{s}=4\pi \mathrm{m}/\mathrm{s} \end{gathered}$$

If the wave equation is written in a different fashion, then also we will get the same values for these quantities.

Answer:

The wave equation is represented by
$y=\left(1·0 \mathrm{mm}\right) \sin \mathrm{\pi }\left(\frac{x}{2·0 \mathrm{cm}}-\frac{t}{0·01 \mathrm{s}}\right)$
Let:
Time period = T
Wavelength = λ

$$\begin{gathered} \left(a\right) T=2\times 0.01=0.02 \mathrm{s}=20 \mathrm{ms} \\ \lambda =2\times 2=4 \mathrm{cm} \end{gathered}$$

(b) Equation for the velocity of the particle:
$\nu =\frac{dy}{dt}=\frac{\mathrm{d}}{dt}\left[\sin 2\mathrm{\pi }\left(\frac{\mathrm{x}}{4}-\frac{t}{0.02}\right)\right]$
$$\begin{gathered} =-0.50 \cos 2\pi \left\{\left(\frac{x}{4}\right)-\left(\frac{t}{0.02}\right)\right\}\times \frac{1}{0.02} \\ \Rightarrow \nu =-0.50 \cos 2\pi \left\{\left(\frac{x}{4}\right)-\left(\frac{t}{0.02}\right)\right\} \\ \mathrm{At} x=1 \mathrm{and} t=0.01 \mathrm{s}, \nu =-0.50 \cos 2\pi \left\{\frac{1}{4}-\frac{1}{2}\right\}=0. \end{gathered}$$

(c) (i) Speed of the particle:
       $\mathrm{At} x=3 \mathrm{cm} \mathrm{and} t=0.01 \mathrm{s},$ $\nu =-0.50\cos 2\pi \left\{\frac{3}{4}-\frac{1}{2}\right\}=0$.
(ii) $\mathrm{At} x=5 \mathrm{cm} \mathrm{and} t=0.01 \mathrm{s},$
$$\begin{gathered} \nu =0 \\ \left(\mathrm{iii}\right) \mathrm{At} x=7 \mathrm{cm} \mathrm{and} t=0.1 \mathrm{s}, \nu =0. \\ \left(\mathrm{iv}\right) \mathrm{At} x=1 \mathrm{cm} \mathrm{and} t=0.011 \mathrm{s}, \\ \nu =50 \cos 2\pi \left\{\left(\frac{1}{4}\right)-\left(\frac{0.011}{0.02}\right)\right\} \\ =-50 \cos \left(\frac{3\pi }{5}\right)=-9.7 \mathrm{cm}/\mathrm{s} \end{gathered}$$
(By changing the value of t, the other two can be calculated.)

Answer:

Time taken to reach from the mean position to the extreme position, $\frac{T}{4}$ = 5 ms
Time period (T) of the wave:
$$\begin{gathered} T=4\times 5 \mathrm{ms} \\ =20\times {10}^{-3}=2\times {10}^{-2} \mathrm{s} \end{gathered}$$
Wavelength (λ) = $2\times \mathrm{Distance} \mathrm{between} \mathrm{two} \mathrm{mean} \mathrm{positions}$
                       $=2\times 2 \mathrm{cm}=4 \mathrm{cm}$
$$\begin{gathered} \mathrm{Frequency}, f=\frac{1}{\mathrm{T}} \\ =\frac{1}{\left(2\times {10}^{-2}\right)} \\ =50 \mathrm{Hz} \\ \mathrm{Wave} \mathrm{speed}, v= \lambda f \\ =4\times {10}^{-2}\times 50 \\ =200\times {10}^{-2} \\ =2 \mathrm{m}/\mathrm{s} \end{gathered}$$

Answer:

Given:
Wave speed, $\nu =20 \mathrm{cm}/\mathrm{s}$
From the graph, we can infer:
(a) Amplitude, A = 1 mm
(b) Wavelength, λ = 4 cm

(c) Wave number, $k=\frac{2\pi }{\lambda }$
                            $$\begin{gathered} =\frac{\left(2\times 3.14\right)}{4} \\ =1.57 {\mathrm{cm}}^{-2} \end{gathered}$$

(d) $\mathrm{Time} \mathrm{period},\mathit{ }T=\frac{\lambda }{\nu }$
$$\begin{gathered} \mathrm{Frequency}, f=\frac{1}{T}=\frac{v}{\lambda } \\ \Rightarrow f=\frac{20}{4}=5 \mathrm{Hz} \end{gathered}$$

Answer:

Given,
Wave speed (v) = 10 ms⁻¹
Time period (T) = 20 ms $=20\times {10}^{-3}=2\times {10}^{-2} \mathrm{s}$

(a) Wavelength of the wave:
$$\begin{gathered} \lambda =\nu t=10\times 2\times {10}^{-2} \\ = 0.02 \mathrm{m}=20 \mathrm{cm} \end{gathered}$$

(b) Displacement of the particle at a certain instant:
$$\begin{gathered} y=a\sin \left(\omega t-kx\right) \\ \Rightarrow 1.5=a\sin \left(\omega t-kx\right) \end{gathered}$$
Phase difference of the particle at a distance x = 10 cm:
$\mathrm{\varphi }=\frac{2\pi x}{\lambda }=2\pi \times \frac{10}{20}=\pi$
$$\begin{gathered} \mathrm{The} \mathrm{displacement} \mathrm{is} \mathrm{given} \mathrm{by} \\ y\text{'}=a\sin \left(\omega t-kx+\pi \right) \\ =a\sin \left(\omega t-kx\right)=1.5 \mathrm{mm} \\ \therefore \mathrm{Displacement}=1.5 \mathrm{mm} \end{gathered}$$

Answer:

Given,
Length of the steel wire = 64 cm
Weight = 5 g
Applied force = 8 N
Thus, we have:
$$\begin{gathered} \mathrm{Mass} \mathrm{per} \mathrm{unit} \mathrm{length}=\frac{5}{64} \mathrm{gm}/\mathrm{cm} \\ \mathrm{Tension}, T=8 \mathrm{N} \\ =8\times {10}^5 \mathrm{dyn} \\ \mathrm{Speed}, v=\sqrt{\left(\frac{T}{\mathrm{m}}\right)} \\ =\sqrt{\frac{\left(8\times {10}^5\times 64\right)}{5}} \\ =3200 \mathrm{cm}/\mathrm{s}=32 \mathrm{m}/\mathrm{s} \end{gathered}$$

Answer:

Given,
Length of the string = 20 cm
Linear mass density of the string = 0.40 g cm⁻¹
Applied tension = 16 N = $16\times {10}^5 \mathrm{dyn}$
Velocity of the wave:
$$\begin{gathered} \nu =\sqrt{\left(\frac{T}{m}\right)} \\ =\sqrt{\frac{\left(16\times {10}^5\right)}{0.4}} \\ =2000 \mathrm{cm}/\mathrm{s} \end{gathered}$$
∴ Time taken to reach the other end $=\frac{20}{2000}=0.01 \mathrm{s}$
Time taken to see the pulse again in the original position$=0.01\times 2=0.02 \mathrm{s}$

(b) At t = 0.01 s, there will be a trough at the right end as it is reflected.

Answer:

Given,
Linear mass density of the string = 0.5 gcm⁻¹
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms⁻¹



The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.
$\mathrm{Phase} \mathrm{change}=0$
(a) $$\begin{gathered} \mathrm{Total} \mathrm{distance}, S=20+20=40 \mathrm{cm} \\ \mathrm{Wave} \mathrm{speed}, \nu =20 \mathrm{m}/\mathrm{s} \end{gathered}$$
Time taken to regain shape:
$\mathrm{Time}=\frac{S}{\nu }=\frac{40}{20}=2 \mathrm{s}$

(b) The wave regain its shape after covering a period distance$=2\times 30=60$ cm
$\therefore \mathrm{Time} \mathrm{period}=\frac{60}{20}=3 \mathrm{s}$

(c) Frequency, $n=\frac{1}{\mathrm{Time} \mathrm{period}}=\frac{1}{3} {\mathrm{s}}^{-1}$
We know:
$n=\frac{1}{2l}\sqrt{\left(\frac{T}{m}\right)}$
Here, T is the tension in the string.
Now,

$$\begin{gathered} m=\mathrm{Mass} \mathrm{per} \mathrm{unit} \mathrm{length} \\ =0.5 \mathrm{gm}/\mathrm{cm} \\ \Rightarrow \frac{1}{3}=\frac{1}{\left(2\times 30\right)} \sqrt{\left(\frac{T}{0.5}\right)} \\ \Rightarrow T=400\times 0.5 \\ =200 \mathrm{dyn} \\ =2\times {10}^{-3} \mathrm{N} \end{gathered}$$

Answer:

Let:
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v₁.
Thus, we have:
​${\nu }_1=\sqrt{\left(\frac{T}{m_1}\right)}$
The mass per unit length can be given as
$$\begin{gathered} m_1=\left(\frac{{\mathrm{\rho }}_1{a}_1{I}_1}{I_1}\right)={\mathrm{\rho }}_1{a}_1 \\ \Rightarrow {\mathrm{\nu }}_1=\sqrt{\left(\frac{T}{{\mathrm{\rho }}_1{a}_1}\right)} ...\left(1\right) \end{gathered}$$
Let the velocity of the first string be v ₂.  
Thus, we have:
$$\begin{gathered} {\nu }_2=\sqrt{\left(\frac{T}{m_2}\right)} \\ \Rightarrow {\nu }_2=\sqrt{\left(\frac{T}{{\rho }_2{a}_2}\right)} ...\left(2\right) \end{gathered}$$
Given,
$$\begin{gathered} {\nu }_1=2{\nu }_2 \\ \Rightarrow \sqrt{\left(\frac{T}{a_1{\mathrm{\rho }}_1}\right)}=2\sqrt{\left(\frac{T}{a_2{\mathrm{\rho }}_2}\right)} \\ \Rightarrow \left(\frac{T}{a_1{\mathrm{\rho }}_1}\right)=4 \left(\frac{T}{a_2{\mathrm{\rho }}_2}\right) \\ \Rightarrow \frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2}=\frac{1}{4} \\ \Rightarrow {\mathrm{\rho }}_1:{\mathrm{\rho }}_2 = 1:4 \end{gathered}$$

Answer:

Given,
Wave equation, $y=\left(0·02 \mathrm{m}\right)\sin \left(1·0 {\mathrm{m}}^{-1}\right)x+\left(30 {\mathrm{s}}^{-1}\right)t$

$$\begin{gathered} \mathrm{Let}: \\ \mathrm{Mass} \mathrm{per} \mathrm{unit} \mathrm{length}, m=1.2\times {10}^{-4} \mathrm{kg}/\mathrm{m} \\ \mathrm{From} \mathrm{the} \mathrm{wave} \mathrm{equation}, \mathrm{we} \mathrm{have}: \\ k=1 {\mathrm{m}}^{-1}=\frac{2\pi }{\lambda } \\ \mathrm{And}, \\ \omega =30 {\mathrm{s}}^{-1}=2\pi f \end{gathered}$$

Velocity of the wave in the stretched string is given by

$$\begin{gathered} \nu =\lambda f=\frac{\omega }{k}=\frac{30}{1} \\ \Rightarrow v=30 \mathrm{m}/\mathrm{s} \\ \mathrm{We} \mathrm{know}: \\ v=\sqrt{\frac{T}{m}} \\ \Rightarrow 30 =\sqrt{\left(\frac{T}{1.2\times {10}^{-4} }\right)} \\ \Rightarrow T=108\times {10}^{-3}=0.108 \mathrm{N} \end{gathered}$$

So, the tension in the string is 0.108 N.

Answer:

Given,
Amplitude of the wave = 1 cm
Frequency of the wave, $f=\frac{200}{2}=100 \mathrm{Hz}$
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N

(a) Velocity of the wave is given by
$v=\sqrt{\frac{T}{m}}$
Thus, we have:
$$\begin{gathered} v=\sqrt{\left(\frac{90}{0.1}\right)}=30 \mathrm{m}/\mathrm{s} \\ \mathrm{Now}, \\ \mathrm{Wavelength}, \lambda =\frac{v}{f}=\frac{30}{100}=0.3 \mathrm{m} \\ \Rightarrow \mathrm{\lambda }=30 \mathrm{cm} \end{gathered}$$

(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
$y=\left(1 \mathrm{cm}\right)\cos 2\pi \left\{\left(\frac{t}{0.01 \mathrm{s}}\right)-\left(\frac{x}{30 \mathrm{cm}}\right)\right\}$    ...(1)

(c) Using $\cos \left(-\theta \right)=\cos \theta$ in equation (1), we get:
$y=1\cos 2\pi \left(\frac{x}{30}-\frac{t}{0.01}\right)$
$$\begin{gathered} \mathrm{Velocity}, v=\frac{dy}{dt} \\ \Rightarrow v=\left(\frac{2\pi }{0.01}\right)\sin 2\pi \left\{\frac{x}{30}-\frac{t}{0.01}\right\} \\ \mathrm{And}, \\ \mathrm{Acceleration}, a=\frac{d\nu }{dt} \\ \Rightarrow a=\left\{\frac{4{\pi }^2}{{\left(0.01\right)}^2}\right\}\cos 2\pi \left\{\left(\frac{x}{30}\right)-\left(\frac{t}{0.01}\right)\right\} \\ \mathrm{When} x=50 \mathrm{cm}, t=10 \mathrm{ms}=10\times {10}^{-3} \mathrm{s}. \end{gathered}$$

Now,

$$\begin{gathered} v=\left(\frac{2\pi }{0.01}\right)\sin 2\pi \left\{\left(\frac{5}{3}\right)-\left(\frac{0.01}{0.01}\right)\right\} \\ =\left(\frac{2\pi }{0.01}\right)\sin \left(2\pi \times \frac{2}{3}\right) \\ =-\left(\frac{2\mathrm{\pi }}{0.01}\right)\sin \frac{4\pi }{3} \\ =- 200\pi \sin \frac{\pi }{3} \\ =-200\pi \times \frac{\sqrt{3}}{2} \\ =-544 \mathrm{cm}/\mathrm{s} \\ =-5.4 \mathrm{m}/\mathrm{s} \end{gathered}$$
In magnitude, v = 5.4 m/s.
Similarly,
$$\begin{gathered} a=\left\{\frac{4{\pi }^2}{{\left(0.01\right)}^2}\right\}\cos 2\pi \left\{\left(\frac{5}{3}\right)-1\right\} \\ =4{\pi }^2\times {10}^4\times \frac{1}{2} \\ \approx 2\times {10}^5 \mathrm{cm}/{\mathrm{s}}^2 \mathrm{or} 2 \mathrm{km}/{\mathrm{s}}^2 \end{gathered}$$

Answer:

Given,
Length of the string, L = 40 cm
Mass of the string = 10 gm
Mass per unit length$=\frac{10}{40}=\frac{1}{4} \left(\mathrm{gm}/\mathrm{cm}\right)$
Spring constant, k = 160 N/m

$$\begin{gathered} \mathrm{Deflection}, x=1 \mathrm{cm} \\ =0.01 \mathrm{m} \\ \mathrm{Tension}, T= kx=160 \times 0.01 \\ \Rightarrow T=1.6 \mathrm{N}=16\times {10}^4 \mathrm{dyn} \\ \mathrm{Now}, \\ v=\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\left(\frac{16\times {10}^4}{{\displaystyle \frac{1}{4}}}\right)} \\ \Rightarrow v=8\times {10}^2 \mathrm{cm}/\mathrm{s}=800 \mathrm{cm}/\mathrm{s} \end{gathered}$$

∴ Time taken by the pulse to reach the spring, $t=\frac{40}{800}=\frac{1}{20}=0.05 \mathrm{s}$

Answer:

Given,
Mass of each block = $m_1=m_2=3.2 \mathrm{kg}$
Linear mass density of wire AB = 10 gm⁻¹ = 0.01 kgm⁻¹
Linear mass density of wire CD = 8 gm⁻¹ = 0.008 kgm⁻¹
For string CD, velocity is defined as $v=\sqrt{\left(\frac{T}{m}\right)}$.
Here, T is the tension and m is the mass per unit length.
For string CD,
$\mathrm{T}=3.2\times g$
Thus, we have:
$$\begin{gathered} v =\sqrt{\frac{\left(3.2\times 10\right)}{0.008}} \\ =\sqrt{\frac{\left(32\times {10}^3\right)}{8}} \\ =2\times 10\sqrt{10} \\ =20\times 3.14\approx 63 \mathrm{ms} \end{gathered}$$

For string AB,
$$\begin{gathered} \mathrm{T}=2\times 3.2g=64 \mathrm{N} \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ v=\sqrt{\left(\frac{\mathrm{T}}{m}\right)} \\ =\sqrt{\left(\frac{64}{0.01}\right)}=\sqrt{6400} \\ =80 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Answer:

Given,
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string:
$$\begin{gathered} m=\frac{4.5\times {10}^{-3}}{2.25} \\ =2\times {10}^{-3} \mathrm{kg}/\mathrm{m} \\ T=2g=20 \mathrm{N} \\ \mathrm{Wave} \mathrm{speed}, \nu =\sqrt{\left(\frac{T}{m}\right)} \\ =\sqrt{\frac{20}{\left(2\times {10}^{-3}\right)}} \\ = \sqrt{{10}^4} \\ = {10}^2 \mathrm{m}/\mathrm{s}=100 \mathrm{m}/\mathrm{s} \end{gathered}$$

Time taken by the disturbance to reach the pulley:
$$\begin{gathered} t=\left(\frac{s}{\nu }\right) \\ =\frac{2}{100}=0.02 \mathrm{s} \end{gathered}$$

Answer:

Given,
Mass of the block = 4.0 kg
Linear mass density, $m=19.2\times {10}^{-3} \mathrm{kg}/\mathrm{m}$
From the free body diagram,
$$\begin{gathered} \mathit{ }T-4g-4a=0 \\ \Rightarrow T=4\left(a+g\right) \\ =4\left(2+10\right)=48 \mathrm{N} \end{gathered}$$



$$\begin{gathered} \mathrm{Wave} \mathrm{speed}, \nu =\sqrt{\left(\frac{T}{m}\right)} \\ =\sqrt{\frac{48}{19.2\times {10}^3}} \\ =\sqrt{\left(2.5\times {10}^{-3}\right)}=50 \mathrm{m}/\mathrm{s} \end{gathered}$$

Answer:

 
Given,
Speed of the transverse pulse when the car is at rest, v₁ = 60 cm s⁻¹
Speed of the transverse pulse when the car accelerates, v₂= 62 cm s⁻¹
Let:
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Now,
$$\begin{gathered} \hspace{0.17em}\mathrm{Wave} \mathrm{speed}, \nu =\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\left(\frac{Mg}{m}\right)} \\ \mathrm{When} \mathrm{car} \mathrm{is} \mathrm{at} \mathrm{rest}: \\ \mathrm{Tension} \mathrm{in} \mathrm{the} \mathrm{string}, T=\mathrm{M}g \\ \Rightarrow v_1=\sqrt{\left(\frac{Mg}{m}\right)} \\ \Rightarrow \frac{Mg}{m}={\left(60\right)}^2 ...\left(\mathrm{i}\right) \end{gathered}$$
When car is having acceleration:
Tension, $T=\sqrt{{\left(Ma\right)}^2+{\left(Mg\right)}^2}$ 
$$\begin{gathered} \mathrm{Again}, {\nu }_2=\sqrt{\left(\frac{T}{m}\right)} \\ \Rightarrow 62=\frac{{\left[{\left(Ma\right)}^2+{\left(Mg\right)}^2\right]}^{1/4}}{m^{1/2}} \end{gathered}$$
$\Rightarrow \sqrt{\frac{\left[{\left(\mathrm{M}a\right)}^2+{\left(\mathrm{M}g\right)}^2\right]}{m}}={\left(62\right)}^2 ...\left(\mathrm{ii}\right)$
From equations (i) and (ii), we get:
$$\begin{gathered} \left(\frac{Mg}{m}\right)\times \frac{m}{\sqrt{{\left(Ma\right)}^2+{\left(Mg\right)}^2}}={\left(\frac{60}{62}\right)}^2 \\ \Rightarrow \frac{g}{\sqrt{\left(a^2+g^2\right)}}=0.936 \\ \Rightarrow \frac{g^2}{\left(a^2+g^2\right)}=0.876 \\ \Rightarrow \left(a^2+100\right) 0.876=100 \\ \Rightarrow a^2=\frac{12.4}{0.867}=14.15 \\ \Rightarrow a=3.76 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Therefore}, \mathrm{acceleration} \mathrm{of} \mathrm{the} \mathrm{car} \mathrm{is} 3.76 \mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$

Answer:

Let,
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity



Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part dθ at angle θ.
So,
Length of the element$=Rd\mathrm{\theta }, \mathrm{mass} =mRd\theta$
Centrifugal force experienced by the element$=\left(mRd\theta \right) {\omega }^{2}R$
Resolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as
$dF=2m{R}^{2}d\theta {\omega }^2 \mathrm{sin\theta }$
$$\begin{gathered} \mathrm{Total} \mathrm{force}, F={\int }_0^{\pi /2}2m{R}^2{\omega }^2\sin \theta d\theta \\ =2m{\mathrm{R}}^2{\omega }^2 \left[-\cos \mathrm{\theta }\right] \\ =2m{\mathrm{R}}^2{\omega }^2 \\ \mathrm{And}, \\ 2T=2m{R}^2{\omega }^2 \\ \Rightarrow T=m{R}^2{\omega }^2 \end{gathered}$$
Velocity of the transverse vibration is given as
$$\begin{gathered} V\text{'}=\sqrt{\left(\frac{T}{m}\right)} \\ V\text{'}=\sqrt{\left(\frac{m{R}^2{\omega }^2}{m}\right)}=\omega R \end{gathered}$$
Linear velocity of the string, V = $\omega R$
∴ Speed of the disturbance, V' =  V

Answer:

(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as
$$\begin{gathered} v=\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\left(\frac{mgx}{m}\right)} \\ \Rightarrow v=\sqrt{\left(gx\right)} \end{gathered}$$

(b) Let the time taken be dt for the small displacement dx.
Thus, we have:

$dt=\frac{dx}{v}=\frac{dx}{\sqrt{\left(gx\right)}}$
$\therefore \mathrm{Total} \mathrm{time}, T=\underset{0}{\overset{L}{\int }}\frac{dx}{\sqrt{\left(gx\right)}}=\sqrt{\left(\frac{4L}{g}\right)}$

(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.

Now,
$$\begin{gathered} t=\underset{0}{\overset{y}{\int }}\frac{dx}{\sqrt{\left(gx\right)}} \\ =\sqrt{\left(\frac{4y}{g}\right)} \end{gathered}$$

∴ Distance travelled by the particle in this time, S = $L-y$
Using the equation of motion, we get:
$$\begin{gathered} \mathrm{S}=ut+\frac{1}{2} g{t}^2 \\ \Rightarrow L-y=\left(\frac{1}{2}\right) g\times \left\{{\left(\sqrt{\frac{4y}{g}}\right)}^2\right\} \\ \Rightarrow L-y=2y \\ \Rightarrow 3y=L \\ \Rightarrow y=\frac{L}{3} \end{gathered}$$
Thus, the particle will meet the pulse at a distance $\frac{\mathrm{L}}{3}$ from the lower end.

Answer:

Given,
Linear density of each of two long strings A and B, m = $1.2\times {10}^{-2} \mathrm{kg}/\mathrm{m}$
String A is stretched by tension T_a= 4.8 N.
String B is stretched by tension T_b= 7.5 N.
Let v_a and v_b be the speeds of the waves in strings A and B.
Now,
$$\begin{gathered} v_{\mathrm{a}}=\sqrt{\frac{T_a}{m}} \\ \Rightarrow v_{\mathrm{a}}=\sqrt{\frac{\left(4.8\right)}{\left(1.2\times {10}^{-2}\right)}}=20 \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{b}}=\sqrt{\frac{T_{\mathrm{b}}}{m}} \\ \Rightarrow {\mathrm{v}}_{\mathrm{b}}=\sqrt{\frac{7.5}{\left(1.2\times {10}^{-2}\right)}}=25 \mathrm{m}/\mathrm{s} \\ {t}_1=0 \mathrm{in} \mathrm{string} \mathrm{A} \\ {t}_2=0+20 \mathrm{ms}=20\times {10}^{-3}=0.02 \mathrm{s} \end{gathered}$$
Distance travelled by the wave in 0.02 s in string A:
s$=20\times 0.02=0.4 \mathrm{m}$
Relative speed between the wave in string A and the wave in string B, v'$=25-20=5 \mathrm{m}/\mathrm{s}$
Time taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m
$t\text{'}=\frac{s}{v\text{'}}=\frac{0.4}{5}=0.08 \mathrm{s}$

Answer:

Given,
Amplitude of the transverse wave, r = 0.5 mm $=0.5\times {10}^{-3} \mathrm{m}$
Frequency, f = 100 Hz
Tension, T = 100 N
Wave speed, v = 100 m/s
Thus, we have:
$$\begin{gathered} \nu =\sqrt{\left(\frac{T}{m}\right)} \\ \Rightarrow {\nu }^2=\left(\frac{T}{m}\right) \\ \Rightarrow m=\frac{T}{{\nu }^2}=\frac{100}{{\left(100\right)}^2} \\ =0.01 \mathrm{kg}/\mathrm{m} \\ \mathrm{Average} \mathrm{power} \mathrm{of} \mathrm{the} \mathrm{source}: \\ {P}_{\mathrm{avg}}=2{\pi }^{2}m\nu r^2{f}^2 \\ =2 {\left(3.14\right)}^2 \left(0.01\right)\times 100\times {\left(0.5\times {10}^{-3}\right)}^2\times \left(100\right) \\ =2\times 9.86\times 0.25\times {10}^{-6}\times {10}^4 \\ =19.7\times 0.0025=0.049 \mathrm{W} \\ =49\times {10}^{-3} \mathrm{W}=49 \mathrm{mW} \end{gathered}$$

Answer:

Given,
Frequency of the wave, f = 200 Hz
Amplitude, A = 1 mm = 10⁻³ m
Linear mass density, m = 6 gm⁻³
Applied tension, T = 60 N
Now,
Let the velocity of the wave be v.
Thus, we have:
$$\begin{gathered} v=\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\frac{\left(60\right)}{\left(6\times {10}^{-3}\right)}} \\ ={10}^2=100 \mathrm{m}/\mathrm{s} \end{gathered}$$

(a) Average power is given as
$$\begin{gathered} {\mathrm{P}}_{average}=2{\pi }^{2}m\nu {\mathrm{A}}^2{f}^2 \\ =2\times {\left(3.14\right)}^2\times \left(6\times {10}^{-3}\right)\times 100\times \left({10}^{-3}\right)\times {200}^2 \\ =473\times {10}^{-3}=0.47 \mathrm{W} \end{gathered}$$

(b) Length of the string = 2 m
Time required to cover this distance:
$$\begin{gathered} t=\frac{2}{100}=0.02 \mathrm{s} \\ \mathrm{Energy}=\mathrm{Power}\times t \\ =0.47\times 0.02 \\ =9.4\times {10}^{-3} \mathrm{J}=9.4 \mathrm{mJ} \end{gathered}$$

Answer:

Given,
Frequency of the tuning fork, f = 440 Hz
Linear mass density, m = 0.01 kgm⁻¹
Applied tension, T = 49 N
Amplitude of the transverse wave produce by the fork = 0.50 mm
Let the wavelength of the wave be $\lambda$.

(a) The speed of the transverse wave is given by
 $\nu =\sqrt{\left(\frac{T}{m}\right)}$
$\Rightarrow v=\sqrt{\frac{49}{0.01}}=70 \mathrm{m}/\mathrm{s}$
$$\begin{gathered} \mathrm{Also}, \\ \nu =\frac{f}{\lambda } \\ \therefore \lambda =\frac{f}{v}=\frac{70}{440}=16 \mathrm{cm} \end{gathered}$$

(b) Maximum speed (v_max) and maximum acceleration (a_max):
$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ y=\mathrm{A} \sin \left(\omega t-kx\right) \end{gathered}$$
$$\begin{gathered} \therefore \nu =\frac{dy}{dt}=\mathrm{A}\omega \cos \left(\omega t-kx\right) \\ \mathrm{Now}, \\ {\nu }_{\max }=\left(\frac{dy}{dt}\right)=\mathrm{A}\omega \\ =0.50\times {10}^{-3}\times 2\pi \times 440 \\ =1.3816 \mathrm{m}/\mathrm{s}. \\ \mathrm{And}, \\ a=\frac{d^{2}y}{d{t}^2} \\ \Rightarrow a=-\mathrm{A}{\omega }^2 \sin \left(\omega t-kx\right) \\ {a}_{\max }=-\mathrm{A}{\omega }^2 \\ =0.50\times {10}^{-3}\times 4{\pi }^2 {\left(440\right)}^2 \\ =3.8 \mathrm{km}/{\mathrm{s}}^2 \end{gathered}$$

(c) Average rate (p) is given by
$$\begin{gathered} p=2{\pi }^2\nu A^2{f}^2 \\ =2\times 10\times 0.01\times 70\times {\left(0.5\times {10}^{-3}\right)}^2\times {\left(440\right)}^2 \\ =0.67 \mathrm{W} \end{gathered}$$