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A source is producing 1500 sound waves in 3 seconds. If the distance covered by a compression and an adjacent rarefaction be 68 cm, find: (a) frequency, (b) wavelength, and (c) velocity, of the sound wave.

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(a) Frequency. We known that frequency of a wave is the number of waves produced in 1 second.
Here, No.of waves produced in 3 seconds = 1500
So, No. of waves produced in 1 second `= (1500)/(3) = 500`
So, the frequency of this sound wave is 500 hertz.
(b) Wavelength. in a sound wave, the distance covered by a compression and an adjacent rarefaction is equal to its wavelength. This distance has been given to be 68 cm. So, the wavelength `(lambda)` of this sound wave is 68 cm.
(c) Velocity. The formula for calculating the velocity of a sound wave is:
`v = f xx lambda`
Here, Frequency `f = 500 Hz` (Calculated above)
And, Wavelength, `lambda = 68cm` (Calculated above) ltbr. `= (68)/(100)m`
`= 0.68m`
Putting these values of f and `lambda` in the above formula, we get:
`v = 500 xx 0.68`
`= 340 m//s`
Thus, the velocity (or speed) of the sound waves is 340 m/s.

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