A string 2 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is

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$320 m / \sec, 120 H z$

$180 m / \sec, 80 H z$

$180 m / \sec, 120 H z$

$320 m / \sec, 80 H z$

answer is D.

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Detailed Solution

$F o r t h i r d h a r m o n i c, f = 3 \frac{v}{2 L} \Rightarrow 3 (\frac{V}{2 l}) = 240 \Rightarrow 3 \times \frac{V}{2 \times 2} = 240 \Rightarrow V = 320 m / \sec F u n d a m e n t a l f r e q u e n c y = \frac{V}{2 l} = \frac{320}{2 \times 2} = 80 H z$

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