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Two wires, each of length 1.2m are stretched between 2 fixed supports. On wire A there is a second harmonic standing wave whose frequency is 660 Hz. However, the same frequency, 660 Hz, is the third harmonic for wire B. Find the speed at which the waves travel on each wire.

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Explanation:

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The speed of the standing waves in wire A and wire B is 792 m/s and 528 m/s respectively.

Frequency modes and velocity:

The frequency of the standing waves formed in a string of length L tied at both ends is given by:

f = nv/2L

where n = 1,2,3,....is the mode of vibration

and v is the speed of the wave

therefore:

v = 2Lf/n

For wire A:

given that, n = 2

and f = 660 Hz

So the speed of the wave is:

v = 2×660×1.2 / 2

v = 792 m/s

For wire B:

given that, n = 3

and f = 660 Hz

So the speed of the wave is:

v = 2×660×1.2 / 3

v = 528 m/s

Learn more about standing waves:

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