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JAMB Physics 2011 Paper

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Question : 44

Total: 44

A length of wire has a frequency of 225Hz when stretched by a force of 225N. If the force increased to 324N, what is the new frequency of vibration?

270 Hz
512 Hz
488 Hz
356 Hz

Solution:

Concept:

For a stretched wire, the frequency of vibration is directly proportional to the square root of the tension.

Explanation:

The relationship is

f \propto \sqrt{T}

, so

f = k\sqrt{T}

, where

k

is a constant.

Given

f_1 = 225\text{ Hz}

and

T_1 = 225\text{ N}

.

New tension is

T_2 = 324\text{ N}

, and we need

f_2

.

Since

k

is constant,

\frac{f_1}{\sqrt{T_1}} = \frac{f_2}{\sqrt{T_2}}

.

Substitute the values:

\frac{225}{\sqrt{225}} = \frac{f_2}{\sqrt{324}}

.

Thus,

f_2 = \frac{225\sqrt{324}}{\sqrt{225}} = \frac{225 \times 18}{15}

.

Calculating gives

f_2 = 270\text{ Hz}

.

Answer:

The new frequency is

270\text{ Hz}

, which is option A.

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