The fundamental frequency of vibration of 1 cm long string is 256 Hz- If the length of the string is reduced to 14cm keeping the tension unaltered- the new fundamental frequency will be

The correct option is C 1024 Hz We know that, n∝1l ∴n_2n_1 = l_1l_2 n_2 = l_1l_2× n_1 = 1× 2561/4 = 1024 Hz . ...

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Standard XII

Physics

String Fixed at Both Ends

The fundament...

Question

The fundamental frequency of vibration of $1 c m$ long string is $256 H z$ . If the length of the string is reduced to $\frac{1}{4} c m$ keeping the tension unaltered, the new fundamental frequency will be

256 H z

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512 H z

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1024 H z

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2048 H z

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Solution

The correct option is C $1024 H z$
We know that, $n \propto \frac{1}{l}$
$∴ \frac{n_{2}}{n_{1}} = \frac{l_{1}}{l_{2}}$
$n_{2} = \frac{l_{1}}{l_{2}} \times n_{1} = \frac{1 \times 256}{1 / 4} = 1024 H z$ .

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The fundamental frequency of vibration of 1 cm long string is 256 Hz. If the length of the string is reduced to 14cm keeping the tension unaltered, the new fundamental frequency will be

The correct option is C 1024 HzWe know that, n∝1l∴n2n1=l1l2n2=l1l2×n1=1×2561/4=1024 Hz.

The fundamental frequency of vibration of $1 c m$ long string is $256 H z$ . If the length of the string is reduced to $\frac{1}{4} c m$ keeping the tension unaltered, the new fundamental frequency will be

The correct option is C $1024 H z$
We know that, $n \propto \frac{1}{l}$
$∴ \frac{n_{2}}{n_{1}} = \frac{l_{1}}{l_{2}}$
$n_{2} = \frac{l_{1}}{l_{2}} \times n_{1} = \frac{1 \times 256}{1 / 4} = 1024 H z$ .