A tuning fork originally in unison with another fork of frequency 260 produces 4 beats per second when a little wax is attached to it. What is the frequency now?
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Given:
A tuning fork originally in unison with another fork of frequency 260 produces 4 beats per second when a little wax is attached to it.
To find:
What is the frequency now?
Solution:
From given, we have,
A tuning fork originally in unison with another fork of frequency 260 produces 4 beats per second when a little wax is attached to it.
The original frequency = f1 = 260 Hz
Number of beats per second = n = 4
Therefore, the new frequency should be less than the original frequency, as the frequency decreases with the attached wax.
The new frequency = f2 = f1 - n
f2 = 260 - 4
∴ The new frequency = 256 Hz
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