Question

if a body of mass 10 kg hanging from a spring)oscillate with your time period of a 6.28 seconds and the spring constant is (G is equal to 10ms Inverse Square)
A) 6.28Nm^1 B.1Nm^1 C. 10Nm^1 D)3.14Nm^1

Answer 1

Time period due to spring is given by $\bold{T=2\pi\sqrt{\frac{m}{k}}}$
Where m is the mass of body attached with spring , k is spring constant and T is the time period of body .

Here, m = 10 Kg , T = 6.28 sec
so, $\bold{6.28=2\pi\sqrt{\frac{10}{k}}}$
Taking square both sides,
(6.28)² = 4π² × 10/k
k = 4π² × 10/(6.28)²
As you know, π = 3.14 {approximately}
so, k = 4 × (3.14) × (3.14) × 10/(6.28 × 6.28) = 10 N/m

Hence, spring constant is 10N/m , e.g., option ( C)