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Question:

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.

Solution:

Velocity of wave in a string is v=√(T/μ) where T=tension in the string and μ=mass per unit length of the string.

here, T=0.5 N and μ=1×10⁻³g/20×10⁻²m = 0.5×10⁻² kg/m

thus, v=√(0.5/0.5×10⁻²)=10 m/s

Wavelength, λ=v/f=10/100=0.1 m=10 cm

∴The separation between successive nodes=λ/2=10/2=5 cm