A particle executing SHM covers a distance equal to half its amplitude in 1 second. The time period of the particle is

Question

A particle executing SHM covers a distance equal to half its amplitude in 1 second. The time period of the particle is (A) 4 s (B) 6 s (C) 8 s (D) 12 s

Tardigrade Admin · Accepted Answer

In SHM , x=A sin ω t=A sin (2 π/T) t At t=1 s , x=(A/2) ∴ (A/2)=A sin (2 π/T) or (1/2)= sin (2 π/T) or sin (π/6)= sin (2 π/T) or T=12 s

Q. A particle executing $S H M$ covers a distance equal to half its amplitude in $1$ second. The time period of the particle is

4 s

6 s

8 s

12 s

In $S H M, x = A sin ω t = A sin \frac{2 π}{T} t$
At $t = 1 s, x = \frac{A}{2}$
$∴ \frac{A}{2} = A sin \frac{2 π}{T}$ or $\frac{1}{2} = sin \frac{2 π}{T}$
or $sin \frac{π}{6} = sin \frac{2 π}{T}$ or $T = 12 s$

Q. A particle executing SHM covers a distance equal to half its amplitude in 1 second. The time period of the particle is

4 s

6 s

8 s

12 s

In SHM,x=Asinωt=AsinT2π​t At t=1s,x=2A​ ∴2A​=AsinT2π​ or 21​=sinT2π​ or sin6π​=sinT2π​ or T=12s

Q. A particle executing $S H M$ covers a distance equal to half its amplitude in $1$ second. The time period of the particle is

In $S H M, x = A sin ω t = A sin \frac{2 π}{T} t$
At $t = 1 s, x = \frac{A}{2}$
$∴ \frac{A}{2} = A sin \frac{2 π}{T}$ or $\frac{1}{2} = sin \frac{2 π}{T}$
or $sin \frac{π}{6} = sin \frac{2 π}{T}$ or $T = 12 s$