First slide

Equation of wave

Question

A string of length 1m has the mass per unit length 0.1 gm/cm . Two ends of the string are attached to two rigid walls. A transverse vibration is set up in the string . A tension of 400 N is maintained in the string 2t the string oscillates in the first overtone and maximum displacement of a particle of the string is 2mm, the maximum velocity of a particle of the string is

Moderate

A

4.371 m/s
B

2.512 m/s
C

3.42 m/s
D

24.8 m/s

Solution

$\begin{array}{l} 2 \times \frac{λ}{2} = l \Rightarrow λ = l = 1 m \\ Velocity of wave, v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{400}{10^{- 2}}} m / s = 200 m / s \\ ∴ Frequency, f = \frac{v}{λ} = \frac{200}{1} Hz = 200 Hz \\ ∴ Maximum particle velocity, u_{\max} = A \times 2 πf = 2 \times 10^{- 3} \times 2 π \times 200 m / s \\ = 2 .512 m / s \end{array}$

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