Tag: speed and acceleration of travelling wave

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Questions Related to speed and acceleration of travelling wave

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 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.

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

The equation of a progressive wave for a wire is: 
$Y=4\sin{\left[\cfrac{\pi}{2}\left(8t-\cfrac{x}{8}\right)\right]}$. If $x$ and $y$ are measured in cm then velocity of wave is :

  1. $64 cm/s$ along $-x$ direction

  2. $32 cm/s$ along $-x$ direction

  3. $32 cm/s$ along $+x$ direction

  4. $64 cm/s$ along $+x$ direction
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} w=4\pi  \ K=\dfrac { \pi  }{ { 16 } }  \ v=\dfrac { w }{ K } =64\, m/s\, along\, \, +x-axis \ Hence, \ option\, \, D\, \, is\, correct\, \, answer. \end{array}$

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

An open tube is in resonance with string (frequency of vibration of tube in $n _{0}$. If tube is dipped on water is that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be 

  1. 1

  2. 2

  3. $\dfrac{2}{3}$

  4. $\dfrac{3}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For open tube no $ = \dfrac{V}{2l} $
For closed tube length available for resonance
$ l^{1} ,l\times \dfrac{25}{100} = \frac{l}{4} $
fundamental frequency of water filled tube 
$ n _{1}\dfrac{V}{4l^{1}} = \frac{V}{4(l/4)} $ $(\because l^{1}= l/4) $
$ \therefore \dfrac{V}{l} = 2n _{0} = 1 $
$ = \dfrac{n}{n _{0}} = 2 $
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The equation of a standing wave in a string fixed at both ends is given as $ y =  A \quad sin \quad  kx \quad cos \quad \omega t $
The amplitude and frequency of a particle vibrating at the mid of an antiode and a node are respectively

  1. $A,\dfrac{\omega }{{2\pi }}$

  2. $\dfrac{A}{{\sqrt 2 }},\dfrac{\omega }{{2\pi }}$

  3. $A,\dfrac{\omega }{{\pi }}$

  4. $\sqrt 2 A,\dfrac{\omega }{{2\pi }}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A wire of length l , area of cross section A  and young's modules of elasticity  y is  suspended from the roof of a building. A  block of mass m is attached at lower end of the wire. if the block is displaced from its mean position and then released the block starts  oscillating. Time period of these oscillation will be

  1. $2\pi \sqrt { \frac { Al }{ mY } } $

  2. $2\pi \sqrt { \frac { AY }{ ml } } $

  3. $2\pi \sqrt { \frac { ml }{ YA } } $

  4. $2\pi \sqrt { \frac { m }{ YAl } } $
Reveal answer Fill a bubble to check yourself
B Correct answer