Tag: wave motion

Quizzes (25) Questions (122)

Questions Related to wave motion

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

A transverse wave propagating on the string can be described by the equation  $y = 2 \sin ( 10 x + 300 t ).$  where $x$  and  $y$  are in metres and  $t$  in second. If the vibrating string has linear density of  $0.6 \times 10 ^ { - 3 } \mathrm { g/cm }$  then the tension in the string is

  1. $5.4 \mathrm { N }$
  2. $0.054 \mathrm { N }$
  3. $54 \mathrm { N }$
  4. $0.0054 N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

General equation of progressive wave:

$y = A \sin ( \omega t + k x )$

Here, $A$ is amplitude

$\omega$ is angular frequency

$k\,$is propagation constant

$t$  is time

$x$ is displacement

Given equation:

$y = 2 \sin ( 10 x + 300 t ).$

Linear charge density, $\mu =0.6\times {{10}^{-3}}\text{g/cm}\,\text{=}\,6\times \text{1}{{\text{0}}^{-3}}\,kg{{m}^{-1}}$
 

Compare with general equation:

Velocity of the wave in the string:

$v = \dfrac { \omega } { k }$$=\dfrac{300}{10}\text{=30}\,\text{m}{{\text{s}}^{-1}}$

Relation for velocity in terms of tension:

$T = \mu v ^ { 2 }$$=6\times {{10}^{-3}}\times {{(30)}^{2}}=5.4\,N$

Hence, tension in string is$5.4\,N$

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

A person observe two points on a string as a travelling wave passes them. The points are at $x _ { 1 } = 0$ and $x _ { 2 } = 1 m.$ The transverse motions of the two points are found to be as follows:
$y _ { 1 } = 0.2 \sin 3 \pi t$
$y _ { 2 } = 0.2 \sin ( 3 \pi t + \pi/8 )$
What is the maximum wavelength?

  1. $32 m$
  2. $16 m$
  3. $8 m$
  4. $4 m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The phase difference delta_phi = k * delta_x. Given y1 = 0.2 sin(3 * pi * t) and y2 = 0.2 sin(3 * pi * t + pi/8), the phase difference is pi/8 for a distance delta_x = 1 m. Thus, k = (pi/8) / 1 = pi/8. Since k = 2 * pi / lambda, we have pi/8 = 2 * pi / lambda, so lambda = 16 m.

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

A string of length $L$ is stretched along the $x-axis$ and is rigidly clamped at its two ends. It undergoes transverse vibration. If $n$ is an integer, which of the following relations may represent the shape of the string at any time:-

  1. $y = A \sin \left( \dfrac { n \pi x } { L } \right) \cos \omega t$
  2. $y = A \sin \left( \dfrac { n \pi x } { L } \right) \sin \omega t $
  3. $y = A \cos \left( \dfrac { n \pi x } { L } \right) \cos \omega t $
  4. $y = A \cos \left( \dfrac { \operatorname { n\pi } x } { L } \right) \sin \omega t$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a string fixed at both ends, the standing wave must satisfy boundary conditions y = 0 at x = 0 and x = L. The form y = A sin(n * pi * x / L) * sin(omega * t) satisfies these conditions.

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

A particle starting from mean position having equation $y=A\sin { \pi t } $ .Find velocity of particle at t=1/3 sec.

  1. $\dfrac { A\pi }{ 2 } $
  2. $\dfrac { \sqrt { 3 } }{ 2 } A\pi $
  3. $A\pi $

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

The equation of motion is y = A sin(pi t). The velocity of the particle is v = dy/dt = A pi cos(pi t). At t = 1/3 seconds, v = A pi cos(pi / 3) = A pi * (1/2) = (A pi) / 2.

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

Two strings of same material are stretched to the same tension. If their radii are in the ratio $1:2$, then respective wave velocities in them will be in ratio

  1. $4:1$
  2. $2:1$
  3. $1:2$
  4. $1:4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We know that the velocity of wave in a stretched string is given by:
$v=\sqrt{\dfrac{TL}{m}}$
Where $T=$tension$=$same for both
$L=$length$=$same for both
$m=$mass
hence
$v\propto \dfrac{1}{\sqrt{m}}$
we know that
mass$=$volume$\times$ density
$=\pi r^2L\rho$
Since L and $\rho$ are equal for both the strings, hence $m\propto r^2$
$\Rightarrow v\propto \dfrac{1}{\sqrt{m}}\propto \dfrac{1}{r}$
$\Rightarrow \dfrac{v _1}{v _2}=\dfrac{r _2}{r _1}=2:1$.
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The equation of a ware is represented by $y = {10^4}\,\sin \,\left[ {100t - \frac{X}{{10}}} \right]$ here $X$ in meter and $t$ in second$.$ The velocity of the wave will be $:-$

  1. $100 m/s$
  2. $250 m/s$
  3. $750 m/s$
  4. $1000 m/s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The wave equation is y = A sin(omega * t - k * x). Here, omega = 100 and k = 1/10. Wave velocity v = omega / k = 100 / (1/10) = 1000 m/s.

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

A stretched string is $1\ m$ long. Its liner density is $0.5\ gm/m$. It is stretched with a force of $20\ N$. If plucked at a distance of $25\ cm$ from one end, the frequency of the tone emitted by it is

  1. $100\ Hz$
  2. $200\ Hz$
  3. $300\ Hz$
  4. $400\ Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The fundamental frequency of a stretched string is given by f = (1 / 2L) * sqrt(T / mu). Given L = 1 m, T = 20 N, and mu = 0.5 g/m = 0.5 * 10^-3 kg/m. Plugging in the values: f = (1 / 2(1)) * sqrt(20 / 0.5 * 10^-3) = 0.5 * sqrt(40000) = 0.5 * 200 = 100 Hz. The plucking point does not affect the fundamental frequency.

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

A particle moves with simple harmonic motion in a straight line. In first $\tau s,$, after starting from rest it travels a distance $a$, and in next $\tau s$ it travels $2a$, in same direction, then:

  1. amplitude of motion is $4a$
  2. time period of oscillations is $6$,
  3. amplitude of motion is $3a$$\tau $
  4. time period of oscillations is $8$,$\tau $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For SHM starting from rest at x = A, x(t) = A cos(omega * t). Distance traveled in time tau is A - A cos(omega * tau) = a. In next tau, distance is A cos(omega * tau) - A cos(2 * omega * tau) = 2a. Solving these equations leads to the amplitude being 4a.