Question

A point mass oscillates along the x-axis according to the law $$x=x_{0}cos\left(\omega t-{\displaystyle \frac{\pi }{4}}\right).$$ If the acceleration of the particle is written as$$a=Acos\left(\omega t+\delta \right)$$, then
A.$$A=x_0,\delta ={\displaystyle \frac{\pi }{4}}$$
B.$$A=x_0{\omega }^2,\delta ={\displaystyle \frac{\pi }{4}}$$
C.$$A=x_0{\omega }^2,\delta =-{\displaystyle \frac{\pi }{4}}$$
D.$$A=x_0{\omega }^2,\delta ={\displaystyle \frac{3\pi }{4}}$$

Answer 1

Hint: Projection of the uniform circular motion on a diameter of the circle in which the circular motion occurs is called simple harmonic motion. Given the displacement of the point mass. We know that velocity can be written as the rate of change of displacement with respect to time. So we need to differentiate this displacement so that we get the velocity of the point mass and then we need to differentiate the velocity in order to get the acceleration. Then we need to equate both the acceleration in order to get our answer.

Complete answer:
Given the displacement of the point mass as, $$x=x_{0}cos\left(\omega t-{\displaystyle \frac{\pi }{4}}\right).$$
We know that velocity $$v$$=$${\displaystyle \frac{dx}{dt}}$$.
Differentiating the displacement, we get$$v={\displaystyle \frac{d}{dt}}\left[x_{0}cos\left(\omega t-{\displaystyle \frac{\pi }{4}}\right)\right]$$
$$v=-x_0\omega \sin \left(\omega t-{\displaystyle \frac{\pi }{4}}\right)$$
The physical significance of the negative sign is that the velocity component of the point is directed to the left, in the negative direction of $$x$$.
And then we need to differentiate the velocity to get the acceleration of the point mass.
$$a={\displaystyle \frac{dv}{dt}}={\displaystyle \frac{d\left[-x_0\omega \sin \left(\omega t-{\displaystyle \frac{\pi }{4}}\right)\right]}{dt}}$$
$$a=-x_0{\omega }^{2}cos(\omega t-{\displaystyle \frac{\pi }{4}})$$
By trigonometric ratios, we know that,
$$\cos (180+\theta )=-\cos \theta$$
Therefore we can write the acceleration as,
$$a=x_0{\omega }^2\left[cos\pi +\left(\omega t-{\displaystyle \frac{\pi }{4}}\right)\right]$$
Simplifying the above equation,
$$a=x_0{\omega }^{2}cos\left(\omega t+{\displaystyle \frac{3\pi }{4}}\right)$$…….. (1)
Also given that
$$a=Acos\left(\omega t+\delta \right)$$ ……… (2)
Equating both equations (1) and (2) we get,
$$A=x_0{\omega }^2$$ and,
$$\delta ={\displaystyle \frac{3\pi }{4}}$$

Therefore the correct option is D.

Note:
Simple harmonic motion can be defined as the motion in which the displacement of the body from its mean position is directly proportional to its restoring force. The restoring force’s direction is always towards the mean position. Simple harmonic motion can also be described as an oscillatory motion. It is said to be a special case of oscillatory motion. All the SHM’s are oscillatory and also periodic motion. But we cannot say that all oscillatory motion is SHM.