Question

The displacement of mean position of particle starting from mean position in shm at ₃ seconds is √3/2 of amplitude its time period will be:

a. 18 sec
b. 6 √3 sec
c. 9 sec
d. 3 √3 sec

Answer 1

The particle's displacement at 3 seconds is √3/2 of the amplitude.

By equating this to the cosine function representing SHM and solving for the angular frequency, we can determine the time period T to be 18 seconds.

Therefore, the correct answer is: option a) 18 sec.

Step-by-step explanation:

A particle executing simple harmonic motion (SHM) has its displacement at a certain time given as a fraction of its amplitude.

Since the displacement after 3 seconds is √3/2 of the amplitude and we know that the cosine function, in its standard form, describes SHM, we can use the relationship:

x(t) = A cos(ωt + φ)

Substituting the given values, we get:

x(3s) = A cos(ω × 3s + φ)

Because we start at the mean position, φ = 0.

The maximum value of the cosine function is 1, which corresponds to the amplitude, while the value at √3/2 of the amplitude corresponds to 60 degrees or π/3 radians. So:

cos(ω × 3s) = √3/2

Thus ω × 3s = π/3 or ω = π/9

The relationship between angular frequency (ω) and period (T) is:

ω = 2π/T

So substituting back, we get:

T = 2π/ω = 2π/(π/9)

= 18s

Hence the correct answer for the time period of the given SHM is 18 seconds, corresponding to option (a).