Use the harmonic frequency relation for a stretched string

For a string fixed at both ends, $f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$ where:

• $L = 1\,\text{m}$
• $\mu = \dfrac{m}{L}$ is the linear mass density
• $n = 1,2,3,\dots$

Find the fundamental frequency from successive harmonics

Two successive harmonics are given as: $750\,\text{Hz} \quad \text{and} \quad 1000\,\text{Hz}$

The difference between successive harmonics is the fundamental frequency: $f_1 = 1000 - 750 = 250\,\text{Hz}$

Find wave speed on the string

Since $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ with $L=1$ m, $250 = \frac{1}{2}\sqrt{\frac{T}{\mu}}$

So, $\sqrt{\frac{T}{\mu}} = 500$

Hence the wave speed is $v = 500\,\text{m/s}$

Compute linear mass density

Given mass of string: $m = 2\times 10^{-5}\,\text{kg}$ and length $L=1$ m, $\mu = \frac{m}{L} = 2\times 10^{-5}\,\text{kg/m}$

Calculate tension

Using $v = \sqrt{\frac{T}{\mu}}$ so $T = \mu v^2$

Substitute values: $T = (2\times 10^{-5})(500)^2$ $T = (2\times 10^{-5})(250000)$ $T = 5\,\text{N}$

Final answer

$\boxed{5}$

Comparison with stored correct answer

Stored correct answer = $5$

Our derived answer matches it.