JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 2)
A string 2.0 m long and fixed at its ends is
driven by a 240 Hz vibrator. The string vibrates
in its third harmonic mode. The speed of the
wave and its fundamental frequency is :-
180m/s, 80 Hz
180m/s, 120 Hz
320m/s, 120 Hz
320m/s, 80 Hz
Explanation
We have:
$$f = {{nv} \over {2l}}$$
$$240 = {{3 \times v} \over {2 \times 2}}$$
$$ \Rightarrow $$ v = 320 m/s
Fundamental frequency = $${v \over {2l}}$$ = 80 Hz.
$$f = {{nv} \over {2l}}$$
$$240 = {{3 \times v} \over {2 \times 2}}$$
$$ \Rightarrow $$ v = 320 m/s
Fundamental frequency = $${v \over {2l}}$$ = 80 Hz.
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