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A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is :
A
1 Hz
B
3 Hz
C
2 Hz
D
4 Hz
Detailed Solution
Maximum speed in SHM: $v_{max} = a\omega = a(2\pi f)$
$\Rightarrow f = \dfrac{v_{max}}{2\pi a}$
Here $v_{max} = 31.4$ cm/s and $a = 5$ cm.
$f = \dfrac{31.4}{2\pi \times 5} = \dfrac{31.4}{10\pi}$
$f = \dfrac{31.4}{31.4} = 1$ Hz
$\Rightarrow f = \dfrac{v_{max}}{2\pi a}$
Here $v_{max} = 31.4$ cm/s and $a = 5$ cm.
$f = \dfrac{31.4}{2\pi \times 5} = \dfrac{31.4}{10\pi}$
$f = \dfrac{31.4}{31.4} = 1$ Hz