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Question:

Two wires W1 and W2 have the same radius r and respective densities ρ₂=4ρ₁. They are joined together at the point O, shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of anti-nodes formed in W1 to W2 is:

1 : 2

4 : 1

1 : 3

1 : 1

Solution:

n₁ = n₂
Linear density (μ) = m/l = ρAl/l = ρA = ρπr²
For W₁: μ₁ = ρ₁πr²
For W₂: μ₂ = ρ₂πr² = 4ρ₁πr²
Velocity of wave (v) = √(T/μ)
v₁ = √(T/μ₁) = √(T/(ρ₁πr²))
v₂ = √(T/μ₂) = √(T/(4ρ₁πr²)) = v₁/2
Frequency (f) = v/2l
f₁ = v₁/(2l₁) = n₁f
f₂ = v₂/(2l₂) = n₂f
Since the point O is midway between the two bridges, l₁ = l₂ = l/2
n₁f = v₁/l and n₂f = v₂/l
n₁/n₂ = v₁/v₂ = 2/1 = 2
Therefore, the ratio of the number of anti-nodes formed in W₁ to W₂ is 1:2