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

Two vibrating strings of the same material but length $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. Both the string vibrate in their fundamental modes, the one of length $L$ with frequency ${v} _{1}$ and other with frequency ${v} _{2}$. The ratio ${v} _{1}/{v} _{2}$ is given by

  1. $2$

  2. $4$

  3. $8$

  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For vibrating string 
frequency $\alpha \dfrac{1}{lenght} \sqrt{\dfrac{Tension}{mass\ per\ unit\ lenght}}$ ---- $(1)$
Since area of first wire is $\dfrac{\pi (2)^{2}}{\pi 1^{2}} = 4$ times pf second wire
its mass per unit length is also $4$ times 
i.e. $\mu _{1}= 4 \mu$ $\mu _{2}= \mu$
$v _{1} = \dfrac{1}{L} \sqrt{\dfrac{T}{4\mu}}= \dfrac{1}{\mu} \sqrt{\dfrac{T}{\mu}}$
$v _{2}= \dfrac{1}{2} \sqrt{\dfrac{T}{\mu}}$
$\dfrac{v _{1}}{v _{2}}= \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}} \times  \dfrac{2l}{1} \sqrt{\dfrac{\mu}{T}}$
$=1$
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A sonometre wire resonates with a given tuning forck forming standing waves with five antinodes between the two bridges when a mass of $9kg$is suspended from the wire. When this mass is replaced by mass $M$, the wire resonates with the same positions of the bridges. Then find the value of square roof of $M$.

  1. $5$

  2. $10$

  3. $25$

  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of vibration of a string 

$n=\dfrac{p}{2l}\sqrt{\dfrac{T}{m}}$
Also number of loops = Number of antinodes.
Hence with 5 antinodes and hanging mass of 9 kg. we have p=5 and T=9g
So,
$n _1=\dfrac{5}{2l}\sqrt{\dfrac{9g}{m}}$
With 3 antinodes and hanging mass M we have p=3 and T=Mg so,
$n _2=\dfrac{3}{2l}\sqrt{\dfrac{Mg}{m}}$
$\because n _1=n _2$
$\dfrac{5}{2l}\sqrt{\dfrac{9g}{m}}=\dfrac{3}{2l}\sqrt{\dfrac{Mg}{m}}$
Squaring both side we get
$25\times9=9\times M$
$M=25\ kg$

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

A $12m$ long vibrating string has the speed of wave $48 m/s$ to what frequency it will resonate?

  1. $2cps$

  2. $4cps$

  3. $6cps$

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given, $S=48m/s,l=12m$

So, Equation of the fundamental frequency:

$v\dfrac{v}{2l}=\dfrac{48}{2\times12}=2cps$

The string will resonate at fundamental frequency as well as first overtone second overtone and so on.
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A travelling wave tube is given by
$y = \dfrac{0.8}{(3x^2 + 12 xt + 12t^2 + 4)}$, where x and y are in m and t is in s . The velocity of the wave

  1. 3 m/s

  2. 5 m/s

  3. 2 m/s

  4. 7 m/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} y=\dfrac { { 0.8 } }{ { 3{ x^{ 2 } }+12xt+12{ t^{ 2 } }+4 } }  \ =\dfrac { { 0.8 } }{ { 3\left( { { x^{ 2 } }+4x+4{ t^{ 2 } } } \right) +4 } }  \ =\dfrac { { 0.8 } }{ { 3{ { \left( { x+2t } \right)  }^{ 2 } }+4 } }  \ =\dfrac { { 0.8 } }{ { 3\times 4{ { \left( { \dfrac { x }{ 2 } +t } \right)  }^{ 2 } }+4 } }  \ \therefore Velocity=2m/s \end{array}$

$\therefore $ Option $C$ is correct.

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

Two sinusoidal waves with same wavelengths and amplitudes travel in opposite directions along a string with a speed $10$ m $s^{-1}$. If the minimum time interval between two instant when the string is flat is $0.5$s, the wavelength of the waves is?

  1. $25$ m

  2. $20$ m

  3. $15$ m

  4. $10$ m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given frequency $f$=$\dfrac { 1 }{ t } $ and velocity $\nu$=10 m/s

We know $\nu =\lambda f\ \lambda =\dfrac { \nu  }{ f } =\dfrac { 10 }{ \frac { 1 }{ 0.5 }  } =5\quad m\ $
Since both the waves are similar but moves in opposite direction its toatl wavelength of the wave will be 10 m

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

Mark out the correct statements with respect to wave speed and particle velocity for a transverse travelling mechanical wave on a string.

  1. The wave speed is same for the entire wave, while particle velocity is different for different points at a particular instant.

  2. Wave speed depends upon property of the medium but not on the wave properties.

  3. Wave speed depends upon both the properties of the medium and on the properties of wave.

  4. Particle velocity depends upon properties of the wave and not on medium properties.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Wave speed v = sqrt(T/mu) depends only on the properties of the medium (tension T and linear density mu). Particle velocity depends on the wave's amplitude and frequency, which are properties of the wave itself.

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

Two waves $Y _ { 1 } =  { a \sin \omega t }$  and  $Y _ { 2 } = \operatorname { asin } ( \omega t + \delta )$  are producing interference, then resultent intensity is:

  1. $a ^ { 2 } \cos ^ { 2 } \delta / 2$

  2. $2 a ^ { 2 } \cos ^ { 2 } \delta / 2$

  3. $3 a ^ { 2 } \cos ^ { 2 } \delta / 2$

  4. $4 a ^ { 2 } \cos ^ { 2 } \delta / 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$Y _{1}+Y _{2}=a\sin\omega t+a\sin(\omega t+8)$
$Y=a\left[\sin\omega t+\sin(\omega t+8)\right]$
$Y=a\left[2\sin\left(\dfrac{\omega t+\omega t+8}{2}\right)\cos\left(\dfrac{8}{2}\right)\right]$
$Y=2a\sin\left(\omega t+\dfrac{8}{2}\right)\cos\left(\dfrac{8}{2}\right)$
$Y=\left[2a\cos \left(\dfrac{8}{2}\right)\right]\sin(\omega t+8/2)$
As Intensity $\alpha A^{2}$
Here $I \alpha \left[2a\cos\left(\dfrac{8}{2}\right)\right]^{2}$
$I\alpha 4a^{2}\cos^{2}\left(\dfrac{8}{2}\right)$
Option $D$ is correct.



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$5.4\,N$