Question

In Melde's experiment, the string vibrates in 4 loops when a 50 g weight is placed in the pan of weight 15 g. To make the strings to vibrates in 6iJoops the weight that has to be removed from the pan is

• A. 0.0007 kg-wt
• B. 0.0021 kg-wt
• C. 0.036 kg-wt
• D. 0.0029 kg-wt

Answer 1

Answer: (C)

0.036 kg-wt

Answer 2

The transverse vibrations of a string are determined by Melde's method.
The frequency of vibration of a string of length l, mass per unit length m and vibrating in p loops under tension T is given by

\hspace40mm n_= \frac{P}{2l} \sqrt{\frac{T}{m}}

or \hspace30mm p \sqrt T = constant
If n, l and m are constant.
Hence, \hspace30mm T \propto \frac{1}{p^2}
\therefore \hspace30mm \frac{T_1}{T_2} = \frac{p^2_2}{p^2_1}

or \hspace20mm \frac{(50+15)}{T_2} = \frac{(6)^2}{(4)^2}

or \hspace30mm \frac{65}{T_2} = \frac{36}{16}

\therefore\hspace30mm T_2 = \frac{65 \times 16 }{36} = 29g

So, weight removed from the pan
\hspace45mm = 65 - 29 = 36g
\hspace45mm = 0.036\, kg-wt