Tag: wave motion

Questions Related to wave motion

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

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

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

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

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

  4. $-\sqrt { 3 } { \pi }$

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Correct Option: A

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

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

  1. 126 cm

  2. 160 cm

  3. 40 cm

  4. 80 cm

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Correct Option: A
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) $.

A stretched wire emits a fundamental note of $256 Hz$. Keeping the stretching force constant and reducing the length of wire by $10 cm$, the frequency becomes $320 Hz$, the original length of the wire is:

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

  1. $100 cm$

  2. $50 cm$

  3. $400 cm$

  4. $200 cm$

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Correct Option: B
Explanation:

frequency of fundamental note, $\upsilon = \dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$

In first case, $256 = \dfrac{1}{2L} \sqrt{\dfrac{T}{m}} $                ......(i)

In second case, $320 =\dfrac{1}{2(L - 10)} \sqrt{\dfrac{T}{m}}$    ......(ii)

dividing (ii) and (i) we get,

$\dfrac{320}{256} = \dfrac{2L}{2(L - 10)}$ or $\dfrac{L}{L - 10} = \dfrac{5}{4}$


$\therefore L = 50 cm$

A uniform wire of length 20 m and weighing 5 kg hangs vertically. If g=10 $ms^{-2}$, then the speed of transverse waves in the middle of the wire is

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

  1. $10 ms ^{-1}$

  2. $10\sqrt2 ms ^{-1}$

  3. $15ms ^{-1}$

  4. $2 ms ^{-1}$

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Correct Option: A
Explanation:

Given,

$m=5kg$
$l=20m$
$\mu=\dfrac{m}{l}=\dfrac{5}{20}=0.25kg/m$
$g=10m/s^2$
Tension in the middle of the wire, $T=\dfrac{m}{2}g$
$T=\dfrac{5}{2}\times 10=25N$
Velocity, $v=\sqrt{\dfrac{T}{\mu}}$
$T=\sqrt{\dfrac{25}{0.25}}=10m/s$
The correct option is A.

The displacement of particles in a string stretched in the $X-$ direction is represented by $y$. Among the following expressions for $y$, those describing wave motion are:

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

  1. $\cos { Kx } \sin { \omega t }$

  2. $-a\cos { \left( Kx-\omega t \right) }$

  3. $-a\cos { \left( Kx+\omega t \right) }$

  4. $-a\sin { \left( Kx-\omega t \right) }$

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Correct Option: A

A man generates a symmetrical plus in a string by moving his hand up and down. At $t=0$ the point in his hand moves downward. The pulse travels with speed $3 m/s$ on the string & his hands passes $6$ times in eacgh seconds from the mean position. Then the point on the string at a distance $3m$ will reach  its upper extreme first time at time $t=$

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

  1. $1.25 sec.$

  2. $1 sec.$

  3. $\frac{{13}}{{12}}\sec $

  4. None

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Correct Option: D
Explanation:

Firstly, we have to draw the pulse wave (see attached diagram).

If hand passes 6 times from the mean position in one second, then we know that string creates 3 wave lengths (λ) or 3 cycles after 1 second.

That means frequency (f) of the wave is $3 Hz.$

Now we can use below equation to calculate the value of wave length.

$V = f\lambda$ (V = velocity of the wave)

$\lambda =\dfrac{ V}{f}$

  $= \dfrac{(3m/s)}{3} = 1 m$

If $\lambda = 1m$, the point which have 3m distance is located at no.(6) (in the diagram).

to reach its upper extreme ----> have to travel 3λ/4 distance

time to travel $3\lambda = 1$ second

time to travel $\lambda = \dfrac{1}{3}$ seconds

time to travel $\dfrac{3\lambda}{4} = (\dfrac{1}{3}) \times (\dfrac{3}{4})$ seconds

 $= \dfrac{1}{4}$ seconds $= 0.25$ seconds

A wave moving with constant speed on a uniform string passes the point $x = 0$ with amplitude $\displaystyle A _{0}$, angular frequency $\displaystyle \omega _{0}$ and average rate of energy transfer $\displaystyle P _{0}$. As the wave travels down the string it gradually loses energy and at the point x = $\displaystyle l $, the average rate of energy transfer becomes $\displaystyle \dfrac{P _{0}}{2}$. At the point x = $\displaystyle l$, angular frequency and amplitude are respectively

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

  1. $\displaystyle \omega _{0}$ and $A _{0}/\sqrt{2}$

  2. $\displaystyle \omega _{0}/\sqrt{2}$ and $A _{0}$

  3. less than $\displaystyle \omega _{0}$ and $A _{0}$

  4. $\displaystyle \omega _{0}/\sqrt{2}$ and $ A _{0}/\sqrt{2}$

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Correct Option: A

A stationary wave $y=0.4\sin \cfrac{2\pi}{40}x\cos 100\pi t$ is produced in a rod fixed at both end. The minimum possible length of the rod is given by:

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

  1. 10 m

  2. $20\sqrt2m$

  3. 20 m

  4. 28 m

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Correct Option: C

Two strings A and B with $\mu= 2 \ kg/m$ and $\mu= 8 \ kg/m$ respectively are joined in series and kept on a horizontal table with both the ends fixed. The tension in the string is 200 N. If a pulse of  amplitude 1 cm travels in A towards the junction, then find the amplitude of reflected and transmitted pulse. 

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

  1. $A _r=2 A _T=7$

  2. $A _r=\dfrac{-1}{3} A _T=\dfrac{2}{3}$

  3. $A _r=8 A _T=9$

  4. $A _r=3 A _t=4$

Show answer
Correct Option: B
Explanation:

Velocity of wave in string A, ${v _A} = \sqrt {\dfrac{T}{{{\mu _A}}}}  = \sqrt {\frac{{200}}{2}}  = 10\,\,m/s$

Velocity of wave in string B,${v _B} = \sqrt {\dfrac{T}{{{\mu _B}}}}  = \sqrt {\frac{{200}}{8}}  = 5\,\,m/s$
Using $k = \dfrac{w}{v} \Rightarrow {k _A} = 0.1w\,\,and\,{k _B} = 0.2w$
Amplitude of reflected pulse, ${A _B} = \dfrac{{{k _A} - {k _B}}}{{{k _A} + {k _B}}}A = \dfrac{{0.1 - 0.2}}{{0.1 + 0.2}} \times 1 =  - \dfrac{1}{3}$
Amplitude of transmitted pulse,${A _T} = A - \left| {{A _R}} \right| = 1 - \dfrac{1}{3} = \dfrac{2}{3}\,\,cm$

A wave travels on a light string. The equation of the wave is Y = A sin(Kx - $\omega$t + 30$^o$). It is reflected from a heavy string tied to an end of the light string at x = 0. If 64% of the incident energy is reflected the equation of the reflected wave

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

  1. $Y = 0.8 A sin(Kx - \omega \ t + 30^o + 180^o)$

  2. $Y = 0.8 A sin(Kx + \omega \ t + 30^o + 180^o)$

  3. $Y = 0.8 A sin(Kx + \omega \ t - 30^o)$

  4. $Y = 0.8 A sin(Kx + \omega$t + 30^o)$

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Correct Option: C