Question

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

Answer 1

$\huge{\underline{\sf{\red{Answer\leadsto40J}}}}$

Given:

Mass = 2kg

Time period = 4seconds

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$\large{\underline{\underline{\blue{\tt{Formulae}}}}}$

• $T= 2\pi \sqrt\dfrac{m}{k}$
• $PE = \dfrac{1}{2} \omega x^2$

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FBD in the attachment

By the figure we conclude that

$\implies$ $Kx = mg$ ___________(1)

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Solution:

From he formula we get

$\cancel{4}= \cancel{2}\pi \sqrt\dfrac{2}{k}$

$\implies$ $\dfrac{2}{\pi} = \sqrt\dfrac{2}{k}$

Squaring Both the sides

$\dfrac{2}{\pi} ^2 = \dfrac{2}{k}$

$\dfrac{\cancel4}{\pi^2} = \dfrac{\cancel2}{k}$

we know that π² = 10

$k= \dfrac{10}{2}$

$\implies$ k = 5 ______________(2)

Now,

From (1) and (2)

$PE = \dfrac{1}{2} (5) \bigg(\dfrac{mg}{k}\bigg)^2$

$\implies$ $PE = \dfrac{1}{2} {\cancel{5}}\times \dfrac{4 \times \cancel {100}}{\cancel{25}}$

$\implies$ $\dfrac{ 4 \times 20 }{2}$

$\implies$ $\large{\boxed{40J}}}$

Answer 2

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

• m = 2kg
• T = 4 sec

★ Using formula:-

T = 2π √(m/k)

$\implies$ 4 = 2π √(2/k)

$\implies$ 2 = π √(2/k)

$\implies$ 4 = π2 (2/k)

$\implies$ k = 2 π2/4

$\implies$ k = π2/2 = 5 N/m

★ We know that:-

$\implies$ F = mg

$\implies$ mg = kx

$\implies$ x = mg/k

$\implies$ 2 x 10/5

$\implies{\boxed{\text{x = 4}}}$

∴ Potential Energy = (1/2) k x 2

$\implies$ (1/2) × 5 × 16 = 5 × 8

$\implies\large{\boxed{\text{40J}}}$

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