Q141 / 160 · TS EAMCET 2022 (18 Jul Shift 1) · Physics

Two waves of amplitudes $A_{1}$ and $A_{2}$ respectively are superimposed. The ratio between the maximum and minimum intensities of the resultant waves is $9 : 4$ . The value of $\frac{A_{2}}{A_{1}}$ is [Assume $A_{1} > A_{2}$ ]

A
$0.66$
Correct answer
B
$0.20$
C
$0.75$
D
$0.44$

Solution

Intensity of a wave $I$ is related with the amplitude of a wave as,
$I \propto A^{2}$ .
The ratio of maximum to minimum intensity will be,
$\frac{I_{m a x}}{I_{m i n}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2} \Rightarrow \frac{A_{1} + A_{2}}{A_{1} - A_{2}} = \frac{3}{2} \Rightarrow \frac{A_{2}}{A_{1}} = \frac{1}{5} = 0.2$

Hence, the correct answer is option: 'B'.

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Two waves of amplitudes A1 and A2 respectively are superimposed. The ratio between the maximum and minimum intensities of the resultant waves is 9:4. The value of A2 A1 is [Assume A1>A2]

Solution

Relevant Questions

Two waves of amplitudes $A_{1}$ and $A_{2}$ respectively are superimposed. The ratio between the maximum and minimum intensities of the resultant waves is $9 : 4$ . The value of $\frac{A_{2}}{A_{1}}$ is [Assume $A_{1} > A_{2}$ ]