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A turning fork vibrating at `500 Hz` falls from rest accelerates at `10 m//s^(2)`. How far below the point of release is the tuning fork when wave with a frequency of `475 Hz` reach the release point ?A. `16.9 m`B. `16 m`C. `1.69 m`D. `1.6 m` |
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Answer» Correct Answer - A Distance travelled by the tuning fork `= (1)/(2) "gt"^(2) = (1)/(2) xx 10 xx 1.837^(2) = 16.875 m` `= 16.9 m` |
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