Question

A sonometer wire under a tension of 40 N vibrates in unison with a tuning fork of frequency 384 Hz. Find the beat frequency when the tension in the wire is reduced by 0.21 N.

Answer 1

Data: T₁ = 40 N, n₁ = 384 Hz, T₂ = 40 − 0.21 = 39.79 N

`n_1 = 1/(2l) sqrt(T_1/m)` and  `n_2 = 1/(2l) sqrt(T_2/m)`

∴ `n_2/n_1 = sqrt(T_2/T_1)`

∴ `n_2/384 = sqrt(39.79/40)`

= `sqrt0.99475 ≃ sqrt0.9948`

log 0.9948 = `overline(1).9977`

                          `underline(× 1/2)`
                     `overline(1).9989`
log = 384 = + 2.5843
                      2.5832
antilog 2.5832 = 383.0

∴ `n_2 = 384 xx sqrt 0.9948`

= 383.0 Hz

∴ The beat frequency n₁ − n₂

= 384 − 383

= 1 Hz