Question

A body describes S.H.M. in a path 0.12 m long. Its velocity at the centre of the line is 0.12 m/s. Find the period, and magnitude of velocity at a distance √3 x 10⁻² m from the central position.
(Ans : 3.142 s, 0.1149 m/s)

Answer 1

A body describe simple harmonic motion in a path of 12m long. its velocity at the centre of the line is 0.12 m/s
velocity at the centre of line , means velocity at mean position
So, v = $\omega$A
Here A is amplitude .
Amplitude is half of path length
So, A = 0.12/2 = 0.06 m

Now, 0.12 = $\omega$0.06
$\omega$ = 2
2π/T = 2
T = π sec or 3.142 sec

Again, v = $\omega\sqrt{A^2-x^2}$
= $2\times\sqrt{6^2-\sqrt{3}^2}\times10^{-2}$
= 0.1149 m/s

Answer 2

12. A body describes S.H.M. in a path 0.12 m long. Its velocity at the centre of the line is 0.12 m/s. Find the period, and magnitude of velocity at a distance 3 × 10-2 m from the central position.