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Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

Transverse waves on a string have wave speed $12.0$ m/s, amplitude $0.05\  m$ and wavelength $0.4\  m$. The waves travel in the $+ x$ direction and at $t = 0$, the $x = 0$ end of the string has zero displacement and is moving upwards. Find the transverse displacement of a point at x = 0.25 m at time t = 0.15 s.

  1. $-4.54 \  cm$

  2. <span>$-5.54 \ &nbsp;cm$</span>

  3. <span>$-3.54 \ &nbsp;cm$</span>

  4. <span>$-9.54 \ &nbsp;cm$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A wave traveling in +x direction is represented by $y=Asin(\omega t-kx)$

where $\omega=2\pi \nu$
and $k=\dfrac{2\pi}{\lambda}$
We know that speed of wave=$v=\lambda\nu$
Thus here
$\omega=2\pi\times \dfrac{v}{\lambda}=2\pi\times \dfrac{12}{0.4}=60\pi s^{-1}$
and $k=\dfrac{2\pi}{0.4}=5\pi m^{-1}$
Thus wave is $y=(0.05m)sin((60\pi s^{-1})t-(5\pi m^{-1})x)$
Thus the displacement of point at $x=0.25m$ and $t=0.15s$ can be found by putting the values in the equation of wave.
Thus $y(x=0.25m,t=0.15s)=-0.0354m=-3.54cm$

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A long string having a cross-sectional area $0.80 mm^2$ mm2and density, $12.5 g/cc$ is subjected to a tension of $64 N$ along the positive x-axis. One end of this string is attached to a vibrator at $x = 0$ moving in transverse direction at a frequency of $20 Hz$. At $t = 0$, the source is at a maximum displacement $y = 1.0 cm.$ What is the displacement of the particle of the string at $x = 50 cm$ at time $t = 0.05 s$ ?

  1. $0.71 cm $

  2. <span>$0.91 cm $</span>

  3. <span>$0.58 cm $</span>

  4. <span>$0.31 cm $</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Mass per unit length of the string=$\mu=\rho A=0.8\times 10^{-6}\times 12.5\times 10^{3}=0.01kg/m$
Thus speed of the wave=$\sqrt{\dfrac{T}{\mu}}=\sqrt{\dfrac{64}{0.01}}=80m/s$

Amplitude of the wave=A=1cm
$\omega=2\pi\nu=40\pi s^{-1}$
$v=\dfrac{\omega}{k}$
$\implies k=\dfrac{40\pi}{80}=\dfrac{\pi}{2} m^{-1}$
Thus the wave equation is $y=Acos(\omega t-kx)$
$=(1cm)cos[(40\pi s^{-1})t-(\dfrac{\pi}{2}m^{-1})x]$
Thus $y(0.5m, 0.05s)=0.71cm$

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

Three component sinusoidal waves progressing in the same direction along the same path have the same period, but their amplitudes are $A$, $\displaystyle \frac{A}{2}$ and $\displaystyle \frac{A}{3}$ respectively. The phase of the variation at any position $x$ on their path at time $t = 0$ are $0$, $\displaystyle -\frac{\pi}{2}$ and $-\pi$ respectively. Find the amplitude and phase of the resultant wave.

  1. <span>$\displaystyle \frac{5}{6} A$, $\displaystyle -tan^{-1} \left (\frac{3}{4} \right )$</span>

  2. <span>$\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{3}{4} \right )$</span>

  3. <span>$\displaystyle \frac{5}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$</span>

  4. <span>$\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The waves with opposite phases superimpose to give a resultant wave of amplitude, $A-\dfrac{A}{3}=\dfrac{2A}{3}$ with phase angle $0$ at time $t=0$.

This superimposes with wave of amplitude $\dfrac{A}{2}$ in with phase $-\dfrac{\pi}{2}$ at $t=0$.

Hence, the resulting wave has amplitude $\sqrt{(\dfrac{2A}{3})^2+(\dfrac{A}{2})^2}=\dfrac{5}{6}A$
The phase of the resulting wave is $tan^{-1}\dfrac{-\dfrac{A}{2}}{\dfrac{2A}{3}}$$=-tan^{-1}\dfrac{3}{4}$

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

Two wave pulses travel in opposite directions on  a string and approach each other. The shape of one pulse is inverted with respect to the other.

  1. The pulse will collide with each other and vanish

    after collision.

  2. The pulses will reflect each other, that is pulse

    going towards right will finally move towards left

    and vice versa.

  3. The pulses will pass through each other but their

    shapes will be modified.

  4. The pulses will pass through each other without

    any change.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two wave pulses travel on a string and meet, they obey the principle of superposition. They pass through each other without being permanently altered, maintaining their original shapes and velocities after the interaction.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

If 'v' is the velocity of sound in a gas then 'v' is directly proportional to (where M, d and T represents molecular weight of gas, density of gas and its temperature respectively.)

  1. $\sqrt{M}$

  2. $\displaystyle \frac{1}{\sqrt{d}}$

  3. $\sqrt{T}$

  4. Both (2) and (3)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The speed of sound in a gas is given by v = sqrt(gamma * P / d) or v = sqrt(gamma * R * T / M). Since P/d = RT/M, v is proportional to sqrt(T) and inversely proportional to sqrt(M) or sqrt(d). Thus, both options 2 and 3 are correct.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

 Standing waves are generated on string laded with a cylindrical body. If the cylinder immersed in water, the length of the loops changes by a factor of 2.2. The specific gravity of the material of the cylinder is 

  1. 1.11

  2. 2.15

  3. 2.50

  4. 1.26
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The frequency of a vibrating string is proportional to the square root of tension. When immersed in water, the tension changes due to the buoyant force. The ratio of frequencies (or loop lengths) relates to the density of the object and the fluid, leading to the specific gravity calculation.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

In a string the speed of wave is 10 m/s and its frequency is 100 Hz . The value of the phase difference at a distance 2.5 cm will be :

  1. ${ \pi }/{ 2 }$

  2. ${ \pi }/{ 8 }$

  3. ${ 3\pi }/{ 2 }$

  4. ${ 2\pi }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Speed of wave $(v) = 10 m/s$

Frequecy$(\gamma)=100Hz$
Wavelength$(\lambda)=\dfrac{10}{100}=\dfrac{1}{10}ms$
$2\pi$ phase is covered in $\dfrac{1}{10}m$
Hence, at distance of 2.5 m, the phase is $\dfrac{2\pi \times 0.025}{0.1}=\dfrac{\pi}{2}$



Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A string of mass $3$kg is under tension of $400$N. The length of the stretched string is $25$cm. If the transverse jerk is stuck at one end of the string find the velocity?

  1. <span>$25 \pi^2 m s^{-2}$</span>

  2. <span>$-5 \pi^2 m s^{-2}$</span>

  3. <span>$5 \pi^2 m s^{-2}$</span>

  4. <span>$-25 \pi^2 m s^{-2}$</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, A = $5cm = 0.05$m, T = $0.2$s


$\therefore \omega = \dfrac{2 \pi}{T} = \dfrac{2 \pi}{0.2} = 10 \pi rad s^{-1}$

Velocity and acceleration of the particle executing SHM at any displacement xi given by

Velocity, $v = \omega \sqrt{A^2 - x^2}$
and acceleration, $a = - \omega^2 x$
when $x = 5 cm = 0.05$m
$\therefore v = 10 \pi \sqrt{(0.05)^2 - (0.05)^2} = 0 and a = - (10 \pi)^2(0.05) = -5 \pi^2 m s^{-2}$

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A travelling wave travelled in string in +x direction with 2 cm/s, particle at x=0 oscillates according to equation y (in mm) $= 2\sin { \left( \pi t+{ \pi  }/{ 3 } \right)  }$. What will be the slope of the wave at x=3 cm and t=1 s

  1. $-\sqrt { 3 } { \pi }/{ 2 }$

  2. $\tan ^{ -1 }{ \left( -\sqrt { 3 } { \pi }/{ 2 } \right) }$

  3. $-\sqrt { 3 } { \pi }/{ 20 }$

  4. $-\sqrt { 3 } { \pi }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The wave equation is y = 2 sin(pi * t - pi * x / v + pi / 3) where v = 2 cm/s. The slope is the partial derivative dy/dx. Calculating dy/dx at x=3 and t=1 yields the negative value of the derivative component.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The wave-function for a certain standing wave on a string fixed at born ends is y(x, t) = 0.5 sin (0.025$\pi$x) cos 500 t where x and y are in centimeters and t is in seconds The shortest possible length of the string is

  1. 126 cm

  2. 160 cm

  3. 40 cm

  4. 80 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\phi \vec { B } .\vec { dl } ={ \mu  } _{ 0 }{ I } _{ enclosed }$.
Inside the hollow pipe, ${ I } _{ enclosed }=0$.
$\therefore$   $\phi \vec { B } .\vec { dl } =0$
$\Rightarrow$  $B=0$ inside the pipe $\longrightarrow \left( A \right) $.