A rod of length 'L' and negligible mass is suspended by two identical strings A B and C D as shown in the figure. A mass ' M ' is suspended from point ' O ' which is at a distance ' x ' from B . If the frequency of the first harmonic of AB is equal to the frequency of the second harmonic of C D , then the value of ' x ' is

Step-by-step solution

Frequency of Harmenic motion is given by: f = n 2 T m Frequency of first Harmonic of AB is given by: f _ A = 1 2 T _ A m Frequency of second Harmonic of CD is given by: f _ C = 2 2 T _ C m = 1 T _ C m Since frequencies are equal. aligned & f _ A = f _ C & 1 2 T _ A m = 1 T _ C m & 1 4 ( ~T _ A )= T _ C & T _ A =4 ~T _ C aligned Since the system is in equilibrium the torque about point O is same. aligned & T _ A ( x )= T _ C ( L - x ) & 4 ~T _ C ( x )= T _ C ( L - x ) & 4 x = L - x & 5 x = L & x = L 5 aligned