Question

The shortest distance travelled by a particle performing SHM from its mean position in 2s is $1/\sqrt{(}2)$ of its amplitude. Find its time period:

• A. 8
• B. 16
• C. 4
• D. 6

Answer 1

Answer: (B)

16

Let $x=Asinwt$ be the equation of SHM. The value of x is given a $x=1/\sqrt{(}2)A$. Substituting, we get $sin\omega t=1/\sqrt{(}2)⟹\omega (2)=\pi /4$ or $\omega =\pi /8⟹T=16s$
The correct option is (b)