Question

In case of two strings vibrating transversally, the lengths are as 1:2, the diameters are as 2:1 and the densities
are as 4:1. When stretched by same load, the ratio of the two frequencies will be
A) 1:16
B) 1:4
C) 1:2
D) 1:2​

Answer 1

Answer:

Here, L1/L2 = 1/2, D1/D2 = R1/R2 = 2 and p1/p2 = 4

Also, T1=T2=T

Also, u1 = π(R1)^2L1p1/L1 = π(2R2)^2.(1/2).L2.4p2/(1/2).L2 = 16u2

Now, v1 = [1/(2L1)] √(T/u1) = [2.{1/(2L2)}]√(T/16u2) = (1/2) {1/(2L2)} √(T/u2) = (1/2) v2

Hence, v1/v2 = 1/2

Answer 2

Answer:

1:2

Explanation:

Given

the lengths are 1:2,

the diameters are 2:1

the densities are as 4:1

Find

The ratio of two frequencies when stretched by the same load

Solution

For a string, $f=\frac{1}{D l} \sqrt{\frac{T}{\pi d}} \mathrm{~T}$ is same

$\frac{f_{1}}{f_{2}}=\frac{D_{1}}{D_{2}} \frac{l_{2}}{l_{1}} \sqrt{\frac{d_{2}}{d_{1}}}=\left(\frac{2}{1}\right) \sqrt{\frac{1}{4}}=\frac{4}{2}=\frac{2}{1}, \frac{f_{2}}{f_{1}}=\frac{1}{2}$

1:2

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