A `20 cm` long string, having a mass of `1.0g`, is fixed at both the ends. The tension in the string is `0.5 N`. The string is into vibrations using an external vibrator of frequency `100 Hz`. Find the separation (in cm) between the successive nodes on the string.

Options

A. 22 cm
B. 5 cm
C. 15 cm
D. 25 cm

Answer

B

Explanation

Stationary waves are produced on the string.
The frequency of the standing wave on the string ( similar to the sonometer ) is `n = P/(2L) sqrt(T/m)` where P is the number of loops. It is given that ` n = 100` Hz.
`T = 0.05 N, m = (10^(-3) kg)/(0.2 m) " and " l = 0.2 ` m
` :. n = P/(2 xx 0.2) sqrt((0.5 xx 0.2)/10^(-3))`
`100 = P/(0.4) sqrt(10^(3) xx 10^(-1))`
` 100 = (P xx 10 xx 10)/4 :. P = 4 `loops
` :. ` Length of one loop ` = 20/4 = 5` cm
` :. ` Distance between two consecutive nodes = 5 cm