Question

A string is stretched between fixed points separated by 40 cm. It is observed to have consecutive resonant frequencies of 340 Hz and 510 Hz. Then, find the speed of wave on the string.

- 144 m/s
- 128 m/s
- 170 m/s
- 136 m/s

Solution

The correct option is **D** 136 m/s

The harmonics of the wave of nth order are given by,

fn=nv2L

Assume that 340 Hz and 510 Hz are nth and (n+1)th order harmonics, as they were consecutive.

∴510−340=(n+1)v2L−nv2L=v2L

⇒v=170×2L=170×2×0.4

⇒v=136 m/s

The harmonics of the wave of nth order are given by,

fn=nv2L

Assume that 340 Hz and 510 Hz are nth and (n+1)th order harmonics, as they were consecutive.

∴510−340=(n+1)v2L−nv2L=v2L

⇒v=170×2L=170×2×0.4

⇒v=136 m/s

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