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Question: In a ripple tank when one pulse is sent every tenth of a second, the distance between consecutive pu...

In a ripple tank when one pulse is sent every tenth of a second, the distance between consecutive pulses is30mm. In the same depth of water pulses are produced at half second intervals. What is the new distance between consecutive pulses?

A.0.67mm B.6.0mm C.150mm D.600mm A. 0.67mm \\\ B. 6.0mm \\\ C. 150mm \\\ D. 600mm \\\
Explanation

Solution

The frequency and wavelength of any wave (in this case, the ripple)\left( {in{\text{ }}this{\text{ }}case,{\text{ }}the{\text{ }}ripple} \right) gives the speed of the wave,
v=fλv = f\lambda
Under the same depth (that means essentially the same density and pressure, as well as temperature), the velocity of the wave will remain the same.
vinitial=vfinal{v_{initial}} = {v_{final}}

Complete step by step answer:
The wave speed in a medium depends on the frequency and the wavelength of the wave. We are given that initially, one pulse is sent every tenth of a second, the distance between consecutive pulses is30mm. In the second case, at the same depth of water pulses are produced at half second intervals, we need to find out the distance between the consecutive pulses, that is, the wavelength of the wave.

From above, we’ve.

f1=1T1=11/10=10s1 λ1=30mm f2=1T2=11/2=2s1  {f_1} = \dfrac{1}{{{T_1}}} = \dfrac{1}{{1/10}} = 10{s^{ - 1}} \\\ {\lambda _1} = 30mm \\\ {f_2} = \dfrac{1}{{{T_2}}} = \dfrac{1}{{1/2}} = 2{s^{ - 1}} \\\

As we know that, at the same depth under water velocity of sound will be same,
vinitial=vfinal{v_{initial}} = {v_{final}}
=>f1λ1=f2λ2= > {f_1}{\lambda _1} = {f_2}{\lambda _2}
=>λ2=f1λ1/f2= > {\lambda _2} = {f_1}{\lambda _1}/{f_2}
Putting the given values in the above equation we’ve,

=>λ2=10×302 =>λ2=150  = > {\lambda _2} = \dfrac{{10 \times 30}}{2} \\\ = > {\lambda _2} = 150 \\\

So, the distance between consecutive ripples is found to be 150 mm150{\text{ }}mm when the pulses are created at half second interval

So, the correct answer is “Option C”.

Note:
The wave works in such a way that the increase in frequency (decrease in time period)\left( {decrease{\text{ }}in{\text{ }}time{\text{ }}period} \right) and increase in wavelength will lead to increase in wave speed.
Notice that the wavelength has increased from 30mm to 150 mm30mm{\text{ }}to{\text{ }}150{\text{ }}mm due to decrease in frequency from 10 Hz to 2 Hz10{\text{ }}Hz{\text{ }}to{\text{ }}2{\text{ }}Hz, as the wave speed has to remain constant.