Question

The shortest distance travelled by a particle executing SHM from the mean position in 2s is equal to √3/2 times its amplitude, determine its time period.​

Answer 1

Answer:

12

Explanation:

Given  

The shortest distance travelled by a particle executing SHM from the mean position in 2s is equal to √3/2 times its amplitude, determine its time period.

ANSWER

Displacement of a particle performing simple harmonic motion  is given by

                  x = a sin (ωt + ꭤ)

              a √3 / 2 = a sin (ωt + 0)

                  √3 / 2 = sin ωt

            ωt = sin^-1(√3 / 2) = π / 3

            (2π / T) t = π / 3

             T = 2 x 2 x 3 = 12 secs

So time period is 12 secs

Answer 2

Explanation:

Let equation of SHM be:

x=Asin(ωt)

From the given data,

√3A/2=Asin(2ω)

2ω=π/3

2×2π/T=π/3

T=12 s