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 6 Joops the weight that has to be removed from the pan is

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 6 Joops 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

Tardigrade Admin · Accepted Answer

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

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

or

p T

= constant
If

n, l

and m are constant.
Hence,

T \propto \frac{1}{p ^{2}}

∴ \frac{T _{1}}{T _{2}} = \frac{p _{2}^{2}}{p _{1}^{2}}

or

\frac{( 50 + 15 )}{T _{2}} = \frac{( 6 ) ^{2}}{( 4 ) ^{2}}

or

\frac{65}{T _{2}} = \frac{36}{16}

∴ T_{2} = \frac{65 \times 16}{36} = 29 g

So, weight removed from the pan

= 65 - 29 = 36 g

= 0.036 k g - wt

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

0.0007 kg-wt

0.0021 kg-wt

0.036 kg-wt

0.0029 kg-wt

Q. 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 $6$ Joops the weight that has to be removed from the pan is