A string clamped at both its ends is stretched out, it is then made to vibrate in its fundamental mode at a frequency of 45 Hz. The linear mass density of the string is \( 4.0 × 10^{-2}kg/m\) and its mass is \( 2 × 10^{ -2} kg\). Calculate:

(i) the velocity of a transverse wave on the string,

(ii) the tension in the string.

Asked by Pragya Singh | 1 year ago |  107

1 Answer

Solution :-

Mass of the string, m = 2 x 10-2 kg
Linear density of the string = 4 x 10-2 kg
Frequency, vF = 45 Hz

We know, length of the wire = m/µ
= \( \dfrac{(2 \times 10^{-2})}{(4 \times10^{-2})}\) = 0.5 m

We know, λ = \( \dfrac{2I}{n}\)
Where, n = number of nodes in the wire.

For fundamental node, n =1
=> λ = 2l
= 2 x 0.5 = 1m

(i) Therefore, speed of the transverse wave, v = λ vF
= 1 x 45 = 45 m/s


(ii) Tension in the string = µ v2
= 4 x10-2 x 45 = 81 N

Answered by Abhisek | 1 year ago

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