A wire under a certain tension, gives a note of fundamental frequency

1.	A wire under a certain tension, gives a note of fundamental frequency 320 Hz. When the tension is changed, the frequency of the fundamental note rises to 480 Hz. Compare the tensions in the wire.
Answer» Data : n₁ = 320 Hz, n₂ = 480 Hz Fundamental frequency is n = \(\frac1{2L}\)\(\sqrt{\frac Tm}\) As the length (L) of the wire and its mass per unit length (m) are kept constant, n ∝ √T ∴ \(\frac{n_1}{n_2}\) = \(\sqrt{\frac{T_1}{T_2}}\) ∴ The ratio of the tensions in the wire, \({\frac{T_1}{T_2}}\) = \(({\frac{n_1}{n_2}})^2\) = \(({\frac{320}{480}})^2\) = \(({\frac{2}{3}})^2\) = \(\frac49\)

A wire under a certain tension, gives a note of fundamental frequency 320 Hz. When the tension is changed, the frequency of the fundamental note rises to 480 Hz. Compare the tensions in the wire.

Answer»

Data : n₁ = 320 Hz, n₂ = 480 Hz Fundamental frequency is n = \(\frac1{2L}\)\(\sqrt{\frac Tm}\)

As the length (L) of the wire and its mass per unit length (m) are kept constant,

n ∝ √T

∴ \(\frac{n_1}{n_2}\) = \(\sqrt{\frac{T_1}{T_2}}\)

∴ The ratio of the tensions in the wire,

\({\frac{T_1}{T_2}}\) = \(({\frac{n_1}{n_2}})^2\) = \(({\frac{320}{480}})^2\) = \(({\frac{2}{3}})^2\) = \(\frac49\)