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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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Answer» (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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