Question

A massless rod bd is suspended by two identical massless string ab and cd of equal lengths. A block of mass m is suspended from point p such that bp is equal to x. If the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of x is

Answer 1

Let T₁ and T₂ be the tension in the string ab and cd respectively.

m= mass of the block that suspended at point p. Let μ be the frequency of oscillation

According to question,

=> (1/2λ)×√(T₁/μ)= (1/λ)×√(T₂/μ)

=> T₂=T₁/4

Let L be the length of rod

Then for rotational equilibrium,

T₁x=T₂(L-x)

T₁x= (T₁/4)(L-x)

x= L/5  

Answer 2

rotate and look into the ans

hope this helps you thank you