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Question

The length of the wire shown in figure between the pulley is 1.5 m and its mass is 12.0 g. find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

A

140 Hz

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B

210 Hz

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C

70 Hz

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D

30 Hz

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Solution

The correct option is C 70 Hz First let's calculate the tension in the string Free body diagram of one of the mass Now for two "loops” to form we know the string is in its 2nd harmonic f2 = 2 × v2L = vL we need to calculate velocity v for that v = √Tμ and μ= masslength = 12 × 10−31.5 Even though the string has mass ,we can assume tension to be uniform because the mass of the string is very less compared to the blocks. v = √90(12 × 10−31.5) = 106.06 ms−1 v2 = 106.061.5 ∼∼ 70 Hz

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