JEE Main 18 March 2021 Shift 1 Solved Paper

Given, the time period of the simple harmonic motion, T=2s
Displacement covered by the particle, x=A∕2
Here, A is the amplitude of the particle.
The general equation of the simple harmonic motion,
x=A‌sin‌ω‌t
⇒‌‌‌

A

2

=A‌sin(‌

2π

T

t)⇒‌

1

2

=sin(‌

2π

2

t)
⇒‌‌sin‌

π

6

=sin(πt)⇒‌

π

6

=πt
⇒‌‌t=‌

1

6

The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is

1

6

s.
Comparing with t=

1

a

Hence, the value of the a is 6 .