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Tems11 [23]
2 years ago
5

A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struc

k at the same time. A small piece of putty is placed on the tuning fork with the known frequency and it's frequency is lowered slightly. When struck at the same time, the two forks now produce a beat frequency of 8 Hz. 1)What is frequency of tuning fork which originally had a frequency of 335 Hz after the putty has been placed on it
Physics
1 answer:
Ganezh [65]2 years ago
3 0

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

or

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x)  = 332.3

x = 2.7 Hz

or

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

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b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

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