A 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 -BHU 1997-A- 64B- 256C- 512D- 1024

The correct option is D 1024 n ∞ 1/l⇒n_2/n_1 = l_1/l_2⇒ n_2 = l_1/l_2 n_1 = 1 × 256/1/4 = 1024 Hz. ...

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Standard VIII

Physics

Characteristics of Sound

A 1 cm long s...

Question

A 1 cm long string vibrates with fundamental frequency of 256 Hz. If the

length is reduced to $\frac{1}{4}$ cm keeping the tension unaltered, the new

fundamental frequency will be

[BHU 1997]

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256

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512

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1024

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Solution

The correct option is D
1024

$n \infty \frac{1}{l} \Rightarrow \frac{n_{2}}{n_{1}} = \frac{l_{1}}{l_{2}} \Rightarrow n_{2} = \frac{l_{1}}{l_{2}} n_{1} = \frac{1 \times 256}{\frac{1}{4}} = 1024 H z .$

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A 1 cm long string vibrates with fundamental frequency of 256 Hz. If the length is reduced to 14 cm keeping the tension unaltered, the new fundamental frequency will be [BHU 1997]

The correct option is D 1024 n ∞ 1l⇒n2n1=l1l2⇒n2=l1l2n1=1×25614=1024Hz.

A 1 cm long string vibrates with fundamental frequency of 256 Hz. If the

length is reduced to $\frac{1}{4}$ cm keeping the tension unaltered, the new

fundamental frequency will be

[BHU 1997]

The correct option is D
1024

$n \infty \frac{1}{l} \Rightarrow \frac{n_{2}}{n_{1}} = \frac{l_{1}}{l_{2}} \Rightarrow n_{2} = \frac{l_{1}}{l_{2}} n_{1} = \frac{1 \times 256}{\frac{1}{4}} = 1024 H z .$