Question

Twi vibrating metal wires of the same material but of lengths 'L' and '2L' have radii '2r' and 'r' respectively. Two wires are vibrating with frequency 'n₁' and 'n₂' respectively. The two wires are stretched under same tension. The ratio of frequencies 'n₁' to 'n₂' is

• A. 8 : 1
• B. 2 : 1
• C. 4 : 1
• D. 1 : 1

Answer 1

Answer: (D)

1 : 1

Answer 2

The frequency of a vibrating string is given by:

$$n = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$

where $\mu = \rho A$ and $A$ is the cross‑sectional area.

For wire 1:

• Length $= L$
• Radius $= 2r \Rightarrow A_1 = \pi (2r)^2 = 4\pi r^2$

$$\mu_1 = \rho (4\pi r^2)$$ $$n_1 = \frac{1}{2L}\sqrt{\frac{T}{4\rho\pi r^2}}$$

For wire 2:

• Length $= 2L$
• Radius $= r \Rightarrow A_2 = \pi r^2$

$$\mu_2 = \rho (\pi r^2)$$ $$n_2 = \frac{1}{2(2L)}\sqrt{\frac{T}{\rho\pi r^2}} = \frac{1}{4L}\sqrt{\frac{T}{\rho\pi r^2}}$$

Now, compute the ratio:

$$\frac{n_1}{n_2} = \frac{\frac{1}{2L}\sqrt{\frac{T}{4\rho\pi r^2}}}{\frac{1}{4L}\sqrt{\frac{T}{\rho\pi r^2}}} = \frac{1}{2L} \times \frac{4L}{1} \times \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1$$

So, the ratio $n_1 : n_2$ is $1:1$.