Question

A particle executing simple harmonic motion of amplitude $5cm$ has maximum speed of $31.4cm\,{s^{ - 1}}$ . The frequency of its oscillation is:
(A) $1Hz$
(B) $3Hz$
(C) $2Hz$
(D) $4Hz$

Answer 1

Hint: In order to answer this question, first we will rewrite the given value of amplitude and the maximum speed with its symbol. And then we will apply the formula of maximum speed in terms of its amplitude and angular velocity (as a particle is oscillating). As angular velocity is directly proportional to the frequency.


Complete step by step solution:
Given that-
Amplitude of a particle when executing S.H.M, $a = 5cm$ .
And, the maximum speed of a particle, ${V_{\max }} = 31.4cm\,{s^{ - 1}}$ .
To find the frequency of its oscillation, we will apply the formula of maximum speed in terms of its amplitude and angular velocity (as a particle is oscillating):
$\therefore {V_{\max }} = a.w$ ……..(i)
where, $a$ is the amplitude of the particle,
$w$ is the angular velocity of the particle,
and as we know that angular velocity is directly proportional to the frequency.
The angular velocity in terms of frequency is, $w = 2\pi f$
where, $f$ is the frequency of its oscillation.
Now, we will put $w = 2\pi f$ in eq(i):-
$ \Rightarrow {V_{\max }} = a.2\pi f$
$
   \Rightarrow f = \dfrac{{{V_{\max }}}}{{2\pi a}} \\
   \Rightarrow f = \dfrac{{31.4}}{{2\pi \times 5}} = \dfrac{{31.4}}{{10\pi }} = 1Hz \\
 $
Therefore, the frequency of the oscillation of a particle is $1Hz$ .
Hence, the correct option is (A) $1Hz$ .

Note:
There are other circumstances to find the frequency of the oscillation. The time period can be used to compute the frequency of oscillation. The time period is the amount of time it takes for one oscillation to occur. The number of oscillations per second is the frequency of the oscillation. If the particle completes one oscillation in \[T{\text{ }}seconds\] , the number of oscillations per second $(f)$ is as follows:
$f = \dfrac{1}{T}$ .