Question

The maximum speed of a particle executing $SHM$ is $1m/s$ and maximum acceleration is $1.57m/s^2$. Its frequency is:
(A). $0.25s^{-1}$
(B). $2s^{-1}$
(C). $1.57s^{-1}$
(D). $2.57s^{-1}$

Answer 1

Hint- When the body is doing simple harmonic motion then the maximum velocity achieved by that particular body is the amplitude times the angular frequency ($v_{max}=\omega A$) and the maximum acceleration is the amplitude times square of angular frequency$(a_{max}={\omega }^{2}A)$.

Complete step-by-step answer:
Here the value of the maximum velocity of the body doing simple harmonic motion is given,
So $v_{max}=\omega A=1m/s$
$\Rightarrow \omega A=1$----equation (1)
Similarly, the value of the maximum acceleration of the body doing simple harmonic motion is also given,
So, $(a_{max}={\omega }^{2}A)$ $=1.57m/s^2$
$\Rightarrow {\omega }^{2}A=1.57$---equation (2)
Finally, we have to calculate the value of the frequency, so for that if we get the value of the angular frequency then ultimately, we will get the value of the frequency. So, with the help of these two equations we will get the value of the frequency.
So now dividing the equation (2) by the equation (1), we get
${\displaystyle \frac{{\omega }^{2}A}{\omega A}}={\displaystyle \frac{1.57}{1}}$
$\Rightarrow {\displaystyle \frac{{\omega }^{2}A}{\omega A}}=1.57$
$\Rightarrow \omega =1.57rad/s$
 Now we know the relation between the angular frequency and the time period,
$\omega ={\displaystyle \frac{2\pi }{T}}$, where $T$ is the time period.
So, with the help of the above relation we will get the value of time period for the particle performing SHM.
$T={\displaystyle \frac{2\pi }{\omega }}$
$\Rightarrow T={\displaystyle \frac{2\pi }{1.57}}$
$\Rightarrow T={\displaystyle \frac{2\times 3.14}{1.57}}$
$\Rightarrow T=4\sec$
Now we know the relation between the time period and the frequency.
$f={\displaystyle \frac{1}{T}}$, where $f$ is the symbol for the frequency.
$\Rightarrow f={\displaystyle \frac{1}{4}}$
$\Rightarrow f=0.25s^{-1}$
Hence the frequency of the body doing simple harmonic motion is $0.25s^{-1}$.
So, the option (A) is the correct answer.

Note- The maximum velocity of a body doing simple harmonic motion occurs at the equilibrium position (x=0) when the mass is moving towards positive amplitude, and the maximum acceleration of the particle occurs at the extreme ends where force is maximum.