Question: A stretched wire emits a fundamental note of 256 Hz. Keeping the stretching force constant and reduc...

Question

A stretched wire emits a fundamental note of 256 Hz. Keeping the stretching force constant and reducing the length of wire by 10 cm, the frequency becomes 320 Hz, the original length of the wire is

Answer 1

Solution

Frequency of fundamental note, $\upsilon = \frac{1}{2L}\sqrt{\frac{T}{m}}$

In first case

$256 = \frac{1}{2L}\sqrt{\frac{T}{m}}$ ….. (i)

In second case

$320 = \frac{1}{2(L - 10)}\sqrt{\frac{T}{m}}$ ….. (ii)

Dividing (ii) by (i) we get

$\frac{320}{256} = \frac{2L}{2(L - 10)}or\frac{L}{L - 10} = \frac{5}{4}$

$\therefore L = 50cm$