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 of 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 $η \times 100$ Hz, where the value of $η$ is

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Solution

$f_{n} = \frac{n v}{2 L}$

(n = 2) $f_{1} = \frac{2 v}{2 L}$

(n = 6) $f_{2} = \frac{6 v}{2 L}$
$\frac{f_{2}}{f_{1}} = \frac{\frac{6 v}{2 L}}{\frac{2 v}{2 L}} = 3$ $f_{2} = 3 f_{1} = 3 \times 100 = 300 H z$
$\Rightarrow η = 3$

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fn=nv2L (n = 2) f1=2v2L (n = 6) f2=6v2L f2f1=6v2L2v2L=3 f2=3f1=3×100=300 Hz ⇒η=3

$f_{n} = \frac{n v}{2 L}$

(n = 2) $f_{1} = \frac{2 v}{2 L}$

(n = 6) $f_{2} = \frac{6 v}{2 L}$
$\frac{f_{2}}{f_{1}} = \frac{\frac{6 v}{2 L}}{\frac{2 v}{2 L}} = 3$ $f_{2} = 3 f_{1} = 3 \times 100 = 300 H z$
$\Rightarrow η = 3$