Question

A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring.

Answer 1

m = 2kg.

T = 4 sec.

T = 2π √(m/k)

⇒ 4 = 2π √(2/k)

⇒ 2 = π √(2/k)

⇒ 4 = π2 (2/k)

⇒ k = 2 π2/4

⇒ k = π2/2 = 5 N/m

But, we know that F = mg

= kx

⇒ x

= mg/k

= 2 x 10/5

= 4

∴Potential Energy = (1/2) k x2

= (1/2) × 5 × 16 = 5 × 8

= 40J

Answer 2

Given conditions, m = 2kg.  T = 4 sec.

Using the formula,

    T = 2π √(m/k)

∴ 4 = 2π √(2/k)

4 = π² × 2/k

k = π²/2

k = 4.93 N/m.

Now, We know, in vertical Equilibrium,

mg = kx

∴ x = mg/k

∴ x = 2 × 10/4.93

∴  x ≈ 4

Now, Using the Formula,

U = 1/2 kx².

∴ U = 1/2 × 4.93 × 4²

∴ U = 40.3 J. ≈ 40 J.

Hope it helps.