Question

A body oscillates with SHM according to the equation (in S.I. unit) $x = \cos [2\pi t + \dfrac{\pi }{4}]$. At $t = 1.5\sec $, calculate the
(1) Displacement
(2) Speed and
(3) Acceleration of the body

Answer 1

Hint: Simple harmonic motion is a special type of periodic motion. In this type of motion the acceleration of the particle is directly proportional to the displacement of the body from its mean position. The force is always directed towards the mean position. It is a special case of oscillatory motion.

Complete step by step solution:
1. Displacement is defined as the change in the position of the object.
Given $x = \cos [2\pi t + \dfrac{\pi }{4}]$---(i)
At $t = 1.5\sec $
Substituting the value of time in the equation (i),
$\Rightarrow {x_{t = 1.5}} = \cos [2\pi \times 1.5 + \dfrac{\pi }{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos [3\pi + \dfrac{\pi }{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos [\dfrac{{7\pi + \pi }}{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos \dfrac{{8\pi }}{4}$
$\Rightarrow {x_{t = 1.5}} = \cos [2\pi ]$
$\cos n\pi = 1$ if n is an even number.
$\therefore $ ${x_{t = 1.5}} = 1m$
Displacement of the particle is $x = 1m$

2. Speed is the rate of change of displacement with time. It can be writ
$ \Rightarrow v = \dfrac{{dx}}{{dt}}$
$\Rightarrow v = \dfrac{d}{{dt}}[\cos (2\pi t + \dfrac{\pi }{4})]$
$\Rightarrow v = - \sin (2\pi t + \dfrac{\pi }{4}).2\pi $---(ii)
At $t = 1.5\sec $
Substituting the value of time in equation (ii)
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (2\pi \times 1.5 + \dfrac{\pi }{4})$
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (3\pi + \dfrac{\pi }{4})$
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (2\pi )$
$\sin n\pi = 0$ if n is an even number
$\therefore {v_{t = 1.5}} = 0$
The speed of the particle is $v = 0$.

3. Acceleration is the rate of change of velocity with respect to time. It can be written by using the formula
$a = \dfrac{{dv}}{{dt}}$
$\Rightarrow a = \dfrac{d}{{dt}}[ - 2\pi \sin (2\pi t + \dfrac{\pi }{4})]$
$\Rightarrow a = - 2\pi \cos (2\pi t + \dfrac{\pi }{4}) \times 2\pi $---(iii)
At $t = 1.5\sec $
Substituting the value of time in equation (iii)
$\Rightarrow a = - 2\pi \cos (2\pi \times 1.5 + \dfrac{\pi }{4}) \times 2\pi $
$\Rightarrow a = - 4{\pi ^2}\cos (3\pi + \dfrac{\pi }{4})$
$\Rightarrow a = - 4{\pi ^2}\cos (2\pi )$
$\cos n\pi = 1$if n is an even integer
$\therefore a = - 4{\pi ^2}$
The acceleration of the particle is $a = - 4{\pi ^2}$.

Note: It is to be noted that though simple harmonic motion is a special case of oscillatory motion, they are different. Oscillatory motion is the to and fro motion of the particle from its mean position. An object can be in oscillatory motion forever in the absence of friction. But simple harmonic motion is a type of oscillatory motion in which the particle moves along a straight line such that magnitude is proportional to the distance.