First slide

Standing waves

Question

A stretched string of 1 m length, fixed at both ends, having a mass of 5 x $10^{- 4}$ kg is under a tension of 20 N. It is plucked at a point situated at 25 cm from one end. The stretched string would vibrate with a frequency of-

Moderate

A

400 Hz
B

100 Hz
C

200 Hz
D

256 Hz

Solution

Here, $ℓ = 1 m, m = 5 \times 10^{- 4} kg, T = 20 N$
We have, $v = \frac{1}{2 ℓ} \sqrt{\frac{T}{μ}} = \frac{1}{2 (1)} \sqrt{\frac{20}{5 \times 10^{- 4}}} = \frac{1}{2} \sqrt{4 \times 10^{4}} = \frac{2 \times 10^{2}}{2} = 10^{2} = 100 Hz$

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