Ncert Exercises & Solutions

The displacement of a particle varies with time according to the relation, \(y=a~\text{sin} \omega t+b~\text{cos} \omega t.\)

1.	the motion is oscillatory but not SHM
2.	the motion is SHM with amplitude \(a+b\)
3.	the motion is SHM with amplitude \(a^{2}+b^{2}\)
4.	the motion is SHM with amplitude \(\sqrt{a^{2}+b^{2}}\)

Hint: Apply the concept of the superposition principle.

Step 1: Find the combined equation of the motion.

According to the question, the displacement,

y=a sin

ω t

+bcos

ω t

Let a=Asin

ϕ

and b=Acos

ϕ

Now,

a^{2} + b^{2} = A^{2} \sin^{2} ϕ + A^{2} \cos^{2} ϕ

= A^{2} \Rightarrow A = \sqrt{a^{2} + b^{2}}

y = A \sin ϕ \sin ω t + A \cos ϕ \cos ω t = A \sin (ω t + ϕ)

Step 2 : Find the acceleration of the motion .

\frac{d y}{d t} = A ω \cos (ω t + ϕ)

\frac{d^{2} y}{d t^{2}} = - A ω^{2} \sin (ω t + ϕ) = - A y ω^{2} = (- A ω^{2}) y

\Rightarrow \frac{d^{2} y}{d t^{2}} \propto (- y)

Therefore, it is an equation of SHM with amplitude

A = \sqrt{a^{2} + b^{2}}

Hence, option (4) is the correct answer.