1A 1 cm long string vibrates with fundamental frequency of 256 Hz- If the length is reduced to 1-4 cm keeping the tension unaltered- the new fundamental frequency will be in Hz string is fixed at both endsA- 64B- 1024C- 256D- 512

Given, l = 1 cm = 0.01 m Fundamental frequency for string will be, f = v2l = v2l = 12l√(Tμ) For the given string 'μ' will be fixed and tension is kept unchange ...

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Byju's Answer

Standard XII

Physics

String Fixed at Both Ends

1A 1 cm long ...

Question

A 1cm long string vibrates with fundamental frequency of 256 Hz. If the length is reduced to $\frac{1}{4} c m$ keeping the tension unaltered, the new fundamental frequency will be (in Hz) (string is fixed at both ends)

64

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256

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512

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1024

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Solution

The correct option is B : 1024
Given Data:
Length of string, $l = 1 c m = 0.01 m$
Fundamental frequency for string will be,
$f = \frac{v}{2 l} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
For the given string $μ$ will be fixed and tension is kept unchanged. So we can say;
$f \propto \frac{1}{l}$
$\frac{f_{1}}{f_{2}} = \frac{l_{2}}{l_{1}} ⟹ f_{2} = f_{1} \times \frac{l_{1}}{l_{2}}$
$f_{2} = 256 \times \frac{l_{1}}{\frac{l_{1}}{4}} = 256 \times 4 = 1024$

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A 1cm long string vibrates with fundamental frequency of 256 Hz. If the length is reduced to 14cm keeping the tension unaltered, the new fundamental frequency will be (in Hz) (string is fixed at both ends)

The correct option is B : 1024Given Data:Length of string,l=1cm=0.01m Fundamental frequency for string will be,f=v2l=12l√Tμ For the given string μ will be fixed and tension is kept unchanged. So we can say;f∝1lf1f2=l2l1⟹f2=f1×l1l2f2=256×l1l14=256×4=1024

A 1cm long string vibrates with fundamental frequency of 256 Hz. If the length is reduced to $\frac{1}{4} c m$ keeping the tension unaltered, the new fundamental frequency will be (in Hz) (string is fixed at both ends)