To increase the frequency of transverse oscillations of a stretched string by 40%, the tension must be increased by

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TestHub · Accepted Answer

The frequency of transverse oscillations of a stretched string is given by $f = \frac{1}{2 L} \frac{T}{μ}$

This implies $f \propto T$

Let the initial frequency be $f_{1}$ and the initial tension be $T_{1}$ . If the frequency is increased by $40%$ , the new frequency is $f_{2} = f_{1} + 0.40 f_{1} = 1.4 f_{1}$

Using the relation $f \propto T$ , we get $\frac{f _{2}}{f _{1}} = \frac{T _{2}}{T _{1}}$

Substituting the values, $1.4 = \frac{T _{2}}{T _{1}}$

Squaring both sides, $\frac{T _{2}}{T _{1}} = 1.96$

The percentage increase in tension is $\frac{T _{2} - T _{1}}{T _{1}} \times 100% = (1.96 - 1) \times 100% = 96%$

Answer: $96%$

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