Question

The string stretched by tension T and length L vibrates with fundamental frequency n. The tension in the stretched string string is increased sed by 69% and length of the string reduces by 35%. Then the frequency of vibrating

Answer 1

2n

Answer 2

The fundamental frequency of a stretched string is given by

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

where $T$ is the tension, $L$ is the length, and $\mu$ is the linear density.

Given changes:

• Tension increased by 69%: $T' = T + 0.69T = 1.69T$.
• Length reduced by 35%: $L' = L - 0.35L = 0.65L$.

New frequency:

$$f' = \frac{1}{2L'}\sqrt{\frac{T'}{\mu}} = \frac{1}{2(0.65L)}\sqrt{\frac{1.69T}{\mu}} = \frac{1}{2L}\cdot\frac{1}{0.65}\sqrt{1.69}\sqrt{\frac{T}{\mu}}.$$

Since $\sqrt{1.69} = 1.3$, we have:

$$f' = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\cdot \frac{1.3}{0.65} = f\cdot 2.$$

Thus, the new frequency is $2f$ (or $2n$ if $n = f$).