Question

In Melde's experiment, tuning fork was arranged in parallel position and 6 loops were formed along a length of 7.2 m of the string stretched by a weight of 10g. If mass of the string is 14.4 x 10⁻² g, find the frequency of tuning fork. (Ans : 58.33 Hz)

Answer 1

n= 6
l=7.2 m
w=10 g
m=14.4*10raise to power -2

Answer 2

Given :

p = 6  

L = 7.2 m

M = 10 g = 10 × 10⁻³ = 10⁻² kg

M′ = 14.4 × 10⁻² g = 14.4 × 10⁻⁵ kg

n = P / 2L√(T/ m)

For parallel position , frequency is given by  

N=2n

N= 2. P / 2L√(T/ m)

=P/L √(T/m)------1

As we know that m= M’/l

M=14.4x10-5/7.2

M=2x10-5 kg/m

Let us substitute this value in equation 1

N=6/7.2√ [ 10-2x9.8/2x10⁻⁶

=6/7.2 √[49x10⁻²]

=6x7x10/7.2

=70/1.2

=58.33Hz

∴ the frequency of tuning fork. is 58.33 Hz