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 string to vibrates in 6 loops 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

The correct option is C 0.036 kg-wt

The transverse vibrations of a string are determined by vibration of a string of length l, mass per unit length m and vibrating in p loops under tension T is given by
$n=\frac{P}{2l}\sqrt{\frac{T}{m}}$
or $p\sqrt{t}=$ constant
If n, l and m are constant.
Hence, $T\propto \frac{1}{p^2}$
$\therefore \frac{T_1}{T_2}=\frac{p_2^2}{P_1^2}$
or $\frac{(50+15)}{T_2}=\frac{(6{)}^2}{(4{)}^2}$
$\frac{65}{T_2}=\frac{36}{16}$
$\therefore T_2=\frac{65\times 16}{36}=29g$
So, weight removed from the pan
$=65-29$
$=36g$
$=0.036kg-wt$