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Two vibrating strings of the same materi...

Two vibrating strings of the same material but lengths `L` and `2L` have radii `2r` and `r` respectively. They are stretched under the same tension. Both the string vibrate in their fundamental nodes, the one of length `L` with freuqency `v_(1)` and the other with frequency `v_(2)`. the ratio `v_(1)//v_(2)` is given by

A

1

B

`1/2`

C

`3/2`

D

2

Text Solution

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The correct Answer is:
To solve the problem, we need to find the ratio of the frequencies \( v_1 \) and \( v_2 \) of two vibrating strings of different lengths and radii but under the same tension. ### Step-by-Step Solution: 1. **Understanding the Frequency Formula**: The frequency \( v \) of a vibrating string is given by the formula: \[ v = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where: - \( L \) is the length of the string, - \( T \) is the tension in the string, - \( \mu \) is the mass per unit length of the string. 2. **Calculating Mass per Unit Length (\( \mu \))**: The mass per unit length \( \mu \) can be expressed as: \[ \mu = \text{Density} \times \text{Cross-sectional Area} \] For a cylindrical string, the cross-sectional area \( A \) is given by: \[ A = \pi r^2 \] Therefore, for the two strings: - For string 1 (length \( L \), radius \( 2r \)): \[ A_1 = \pi (2r)^2 = 4\pi r^2 \] \[ \mu_1 = \rho \cdot A_1 = \rho \cdot 4\pi r^2 \] - For string 2 (length \( 2L \), radius \( r \)): \[ A_2 = \pi r^2 \] \[ \mu_2 = \rho \cdot A_2 = \rho \cdot \pi r^2 \] 3. **Substituting into the Frequency Formula**: - For string 1: \[ v_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu_1}} = \frac{1}{2L} \sqrt{\frac{T}{4\rho \pi r^2}} = \frac{1}{2L} \cdot \frac{1}{2} \sqrt{\frac{T}{\rho \pi r^2}} = \frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}} \] - For string 2: \[ v_2 = \frac{1}{2(2L)} \sqrt{\frac{T}{\mu_2}} = \frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}} \] 4. **Finding the Ratio of Frequencies**: Now, we can find the ratio \( \frac{v_1}{v_2} \): \[ \frac{v_1}{v_2} = \frac{\frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}}}{\frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}}} = 1 \] ### Final Answer: The ratio \( \frac{v_1}{v_2} \) is: \[ \frac{v_1}{v_2} = 1 \]
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