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A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:A. `1m`B. `75cm`C. `60cm`D. `50cm` |
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Answer» Correct Answer - B Equation of standing wave `y=Asinkxcosomegat` `y=A` as amplitude is 2A `A=2Asinkx` `(2pi)/(lamda)x=(pi)/(6)impliesx_1=(1)/(4)m` and `(2pi)/(lamda).x=(pi)/(2)+(pi)/(3)` `impliesx_1=1.25mimpliesx_2-x_1=1m` |
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