Tag: wave velocity

Questions Related to wave velocity

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. $A _r=3 A _t=4$

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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

What should one do if he wishes to increase the pitch of a string type instrument.
$1$. Increase the length of the string used
$2$. Decrease the gauge of the string used
$3$. Loosen the string
$4$. Tighten the string

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

  1. $1$ and $4$

  2. $2$ and $4$

  3. $2, 1$ and $4$

  4. $3$ and $1$

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Correct Option: B
Explanation:

Thicker, tighter strings, have a more "focussed" sound. They reach their resonant frequency more quickly, because the extra tension leaves them less scope to flap around.

Thicker, tighter strings, plucked the same distance, are louder, because they contain more energy. There is more kinetic energy to be transmitted to the sounding board.


A stretched string is vibrating at $500$ hertz. If the tension is increased four times, the frequency shall become.

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

  1. $1,000$ hertz

  2. $500$ hertz

  3. $250$ hertz

  4. $1,500$ hertz

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Correct Option: A
Explanation:

As frequency $(f)$ is directly proportional to the square root of tension$(t)$.

$f\propto \sqrt t$
$\cfrac{f _1}{f _2}=\sqrt{\cfrac{t _1}{t _2}}$
$\cfrac{500}{f _2}=\sqrt{\cfrac{1}{4}}$
$\implies f=1,000$ hertz