how many vibrations will a tuning fork of frequency 280hz complete during the time sound travels forward in air by 18m?take speed of sound=336m/s
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3
time=336/18=18.667s
no. of vibration =280×18.667=5226.76
no. of vibration =280×18.667=5226.76
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8
Given :
Speed of Sound=V=336m/s
wavelength =18m
Frequency =n=280hz
Speed= distance/time
time= distance/speed=18/336 =0.05sec
Number of vibrations completed in 1 sec=280
Thus, number of vibrations completed in 0.05s
=280x0.05 =14
∴Number of Vibration =14
Speed of Sound=V=336m/s
wavelength =18m
Frequency =n=280hz
Speed= distance/time
time= distance/speed=18/336 =0.05sec
Number of vibrations completed in 1 sec=280
Thus, number of vibrations completed in 0.05s
=280x0.05 =14
∴Number of Vibration =14
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