Question

A long spring whose one end is rigidly fixed is stretched from the other end and then left. Longitudinal waves of frequency 10 Hz are produced. If the velocity of the wave is 25 $m{s}^{-1},$ find the distance between two consecutive compressions in the spring.

• A. 0.5 m
• B. 1.5 m
• C. 2.5 m
• D. 3 m

Answer 1

Answer: (C)

2.5 m

The relationship between the velocity of the sound, its wavelength, and frequency is given by the formula $v=\lambda \times f$,
where,
v - velocity of the wave
$\lambda$- wavelength of the wave
f - frequency of the wave.
In the question, it is given that the frequency is 10 Hz and the velocity of the wave is 25 m/s.
Now the wavelength is calculated from $v=\lambda \times f$ as follows.
That is, $\lambda =\frac{v}{f}$ = $\frac{25m/s}{10Hz}$ = 2.5 m.
Hence, the wavelength of the wave is 2.5 meters