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Q.

A string fixed at both ends is vibrating in the lowest mode of vibration for which a point at quarter of its length from one end is a point of maximum displacement. The frequency vibration in this mode is 100 Hz. The frequency emitted when it vibrates in the next mode such that this point is again a point of maximum displacement is f. The value of f is______

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answer is 300.

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Detailed Solution

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$x = \frac{λ}{4}, \frac{3 λ}{4}, \frac{5 λ}{4}, .....$ are possible values for antinode.
For lowest mode
$\frac{λ}{4} = \frac{l}{4} \Rightarrow λ = l$
Which is as shown above
Also $f = \frac{V}{λ} = \frac{V}{l} = 100 H z$
Now for next higher mode: $\frac{l}{4} = \frac{3 λ_{2}}{4} \Rightarrow λ_{2} = \frac{l}{3}$
$\Rightarrow f_{2} = \frac{V}{λ_{2}} = \frac{V}{l / 3} = \frac{3 V}{l} = 3 \times 100 = 300 H z$

$\Rightarrow$ Ans. 300

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