91Ó°ÊÓ

Two wave pulses are generated in a string. One of the pulses is given by equation \(y_{1}=A \sin (\omega t-k x)\). If average power transmitted by both the pulses along the string are same and is given by \(P=\frac{T A^{2} \omega^{2}}{2 v}\), where \(T\) is the tension in the string, \(A\) is amplitude of a pulse, \(\omega\) is angular frequency of the source, and \(v\) is wave velocity, then which one of the following equations may represent the other wave pulse? (A) \(y_{2}=\frac{A}{\sqrt{2}} \sin (2 \omega t-k x)\) (B) \(y_{2}=\frac{A}{\sqrt{2}} \sin (\omega t-2 k x)\) (C) \(y_{2}=2 A \sin \left(\frac{\omega t}{2}-k x\right)\) (D) \(y_{2}=2 A \sin \left(\frac{\omega t}{2}-\frac{k x}{2}\right)\)

Step by step solution

01

Check Option (A)

Option (A) gives the second wave as \(y_{2} = \frac{A}{\sqrt{2}} \sin(2 \omega t - kx)\). Let's check if it has the same average power as the first wave. Using the power formula: \(P_{2} = \frac{T\left(\frac{A}{\sqrt{2}}\right)^{2}\left(2\omega\right)^{2}}{2v}\), and comparing it to the given power P. If \(P_{2} = P\), then \(TA^2\omega^2 = T\left(\frac{A^2}{2}\right)(4\omega^2)\) This is not true since, \(\frac{1}{2}\neq4\). Therefore, option (A) is incorrect.

02

Check Option (B)

Option (B) gives the second wave as \(y_{2} = \frac{A}{\sqrt{2}} \sin(\omega t - 2kx)\). Calculating the average power for this wave pulse: \(P_{2} = \frac{T\left(\frac{A}{\sqrt{2}}\right)^{2}\omega^{2}}{2v}\), and comparing it to the given power P. If \(P_{2} = P\), then \(TA^2\omega^2 = T\left(\frac{A^2}{2}\right)\omega^2\) This is true since A's coefficients cancel each other. Therefore, option (B) is possible.

03

Check Option (C)

Option (C) gives the second wave as \(y_{2} = 2A \sin\left(\frac{\omega t}{2} - kx\right)\). Calculating the average power for this wave pulse: \(P_{2} = \frac{T(2A)^{2}\left(\frac{\omega}{2}\right)^{2}}{2v}\), and comparing it to the given power P. If \(P_{2} = P\), then \(TA^2\omega^2 = T(4A^2)\left(\frac{\omega^2}{4}\right)\) This is not true since, \(\frac{1}{4}\neq4\). Therefore, option (C) is incorrect.

04

Check Option (D)

Option (D) gives the second wave as \(y_{2} = 2A \sin\left(\frac{\omega t}{2} - \frac{kx}{2}\right)\). Calculating the average power for this wave pulse: \(P_{2} = \frac{T(2A)^{2}\left(\frac{\omega}{2}\right)^{2}}{2v}\), and comparing it to the given power P. If \(P_{2} = P\), then \(TA^2\omega^2 = T(4A^2)\left(\frac{\omega^2}{4}\right)\) This is not true since, \(\frac{1}{4}\neq4\). Therefore, option (D) is incorrect. Having checked all options, we found that the only wave pulse with the same average power transmitted by the first wave pulse is:

05

Answer

The correct equation for the second wave pulse is option (B): \(y_{2} = \frac{A}{\sqrt{2}} \sin(\omega t - 2kx)\).

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