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 vibration using an external vibrator of frequency 100 Hz . Find the separation (in

Correct answer: 5 Distance between the successive nodes, aligned d & =(λ)/(2)=(v)/(2 f) & = T / μ 2 f aligned Substituting the values we get d=5 ~cm

JEE Advanced 2009Paper 2

JEE Advanced 2009 · Paper 2 · Physics · Waves and Sound

A 20 ~cm long string, having a mass of 1.0 ~g , is fixed at both...

20 \mathrm{~cm}

long string, having a mass of

1.0 \mathrm{~g}

, is fixed at both the ends. The tension in the string is

0.5 \mathrm{~N}

. The string is set into vibration using an external vibrator of frequency 100

\mathrm{Hz}

. Find the separation (in

\mathrm{cm}

) between the successive nodes on the string.

Distance between the successive nodes,

\begin{aligned}d & =\frac{\lambda}{2}=\frac{v}{2 f}\\ & =\frac{\sqrt{T / \mu}}{2 f} \end{aligned}

Substituting the values we get d=5

\mathrm{~cm}