Question

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is :

(1) 0.0628 s 

(2) 6.28 s 

(3) 3.14 s 

(4) 0.628 s

Answer 1

Correct option is (4) 0.628 s

For a spring, kx = F

Given, x = 5 cm, F = 10 N

⇒ k(5 × 10⁻²) = 10

⇒ k = \(\frac{1000}5\) = 200 N/m

Now, for spring-mass system undergoing SHM,

\(T = 2\pi \sqrt {\frac mk}\)

Given, m = 2kg

⇒ \(T = 2\pi\sqrt{\frac{20}{200}}\)

\(= \frac{2\pi}{10}\)

= 0.628 s

Answer 2

Option : (4) 0.628 s

F = kx 

10 = k(5 × 10^–2)

k = \(\frac{10}{5\times 10^{-2}}\) 

= 2 x 10²

= 200 N/m

Now,

Answer 3

• Fsp=Kx 
• 10 = k(5 × 10–2)
•  k     =  10/ 5 × 10 − 2
•          =2 x 10² = 200N/m