Question

The maximum speed of a particle executing S.H.M is 1m/s and the maximum acceleration $1.57m/{{s}^{2}}$. Its time period is
a) 4sec
b) 2sec
c) 1.57 sec
d) 1/1.57sec

Answer 1

The maximum speed of a particle performing S.H.M is at the mean position. Whereas the maximum acceleration of the particle is at the extreme position. Hence we will substitute these conditions in the velocity and the acceleration expression of the particle performing S.H.M. Further we will accordingly determine the time period of the particle executing S.H.M.

Formula used:
$v(t)=-\omega \sqrt{{{A}^{2}}-{{x}^{2}}}$
$a(t)=-{{\omega }^{2}}x$
$T=\dfrac{2\pi }{\omega }$

Complete step-by-step answer:
Let us say the above particle executes S.H.M with amplitude (A) i.e. the maximum displacement of the particle with respect to its mean position. Let the angular speed of the particle be $\omega$ and at time ‘t’ let its displacement be ‘x’. Hence the velocity of the particle at time ‘t’ is given by,
$v(t)=-\omega \sqrt{{{A}^{2}}-{{x}^{2}}}$ .
At mean position the velocity of the particle is maximum i.e. at x=0. Hence the above equation becomes,
$\begin{aligned} & v(t)=-\omega \sqrt{{{A}^{2}}-{{x}^{2}}} \\\ & \Rightarrow {{v}_{MAX}}=-\omega \sqrt{{{A}^{2}}-{{0}^{2}}} \\\ & \Rightarrow {{v}_{MAX}}=-\omega A....(1) \\\ \end{aligned}$
Similarly the acceleration of the particle at time ‘t’ is given by,
$a(t)=-{{\omega }^{2}}x$
At extreme position i.e. maximum displacement (A), the acceleration of the particle is maximum. Hence the above equation becomes,
$\begin{aligned} & a(t)=-{{\omega }^{2}}x \\\ & \Rightarrow {{a}_{MAX}}=-{{\omega }^{2}}A...(2) \\\ \end{aligned}$
Taking the ratio of equation 1 and equation 2 we get,
$\begin{aligned} & \dfrac{{{v}_{MAX}}}{{{a}_{MAX}}}=\dfrac{-\omega A}{-{{\omega }^{2}}A} \\\ & \Rightarrow \dfrac{{{v}_{MAX}}}{{{a}_{MAX}}}=\dfrac{1}{\omega } \\\ & \Rightarrow {{a}_{MAX}}=\omega {{v}_{MAX}} \\\ \end{aligned}$
The maximum acceleration as well as the velocity of the particle are given to us. Hence the value of angular velocity is,
$\begin{aligned} & {{a}_{MAX}}=\omega {{v}_{MAX}} \\\ & \Rightarrow 1.57=\omega 1 \\\ & \Rightarrow \omega =1.57rad/s \\\ \end{aligned}$
The relation between time period (T) and the angular velocity is given by,
$T=\dfrac{2\pi }{\omega }$
Hence the time period for the above particle executing S.H.M is,
$\begin{aligned} & T=\dfrac{2\pi }{\omega } \\\ & \Rightarrow T=\dfrac{2\pi }{1.57}=4\sec \\\ \end{aligned}$

So, the correct answer is “Option a)”.

Note: To understand why the particle has maximum velocity at the mean position and the maximum acceleration at the extreme position consider the following example of spring performing S.H.M. At the mean position the mass attached to the spring possesses maximum velocity. At the extreme position the restoring force on the spring is maximum. Since force is proportional to the acceleration we can conclude that the acceleration is also maximum.