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

Answer

5

Explanation

Mass per unit length of the sting :
`mu = (m)/(l) = (10^(-3))/(0.2) = 5 xx 10^(-3)` kg/m
Tension of the string :
F = 0.5 N
Speed of wave can be written as follows :
`upsilon = sqrt((F)/(mu)) = sqrt((0.5)/(5 xx 10^(-3))) = 10` m/s
`upsilon = v lambda rArr 10 = 100 lambda rArr lambda = 0.1 m = 10 cm`
Length of one loop `= lambda//2` = 5 cm
Hence there are four loops in 20 cm length. We can easily understand that there must be five nodes to accommodate four loops. Hence answer is 5.