JEE Main 16 March 2021 Shift 2 Solved Paper

Given,
Mass, m=500g
Decay constant, λ=20g∕s
For damped oscillation,
A=A0e−bbt∕2m....(i)
Here, b= damped constant = decay constant
As per question, the amplitude of the system is dropped to half of its initial value.
So, Eq. (i) becomes
‌

A0

2

=A0e−bt∕2m⇒‌

1

2

=e−bt∕2m
⇒‌‌2=ebt∕2m
⇒‌‌2=ebt∕2m
Taking log on both sides of above equation, we get
log‌2=log(ebt∕2m)
⇒‌‌log‌2=‌

bt

2m

⇒t=‌

2m

b

‌log‌2
=‌

2×500×0.693

20

[∵log‌2=0.693]
⇒t=34.65s