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

$ω t$ +bcos

ω t

$ω t$

Let a=Asin

ϕ

$ϕ$ and b=Acos

ϕ

$ϕ$

Now,

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

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

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

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

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

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

Step 2 : Find the acceleration of the motion .

$Step 2 : Find the acceleration of the motion .$

\frac{d y}{d t} = A ω \cos (ω t + ϕ)

$\frac{d y}{d t} = A ω \cos (ω t + ϕ)$

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

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

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

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

Therefore, it is an equation of SHM with amplitude

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

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

Hence, option (4) is the correct answer.