Question

A body of mass m is attached to the lower in of a spring whose upper end is fixed the spring has negligible mass when the mass m is slightly pulled down and released it oscillates with a time period of 3 second when the mass m is increased by 1 kg the time period of oscillations become 5 second the value of m in kgs​

Answer 1

The mass of the m is m = 9 / 16 Kg

Explanation:

For S.H.M of a hanging mass by a spring:

F = −k  s  x

Where ks  =mω^2

ω =  √  k s / m

Hence time period of S.H.M =T=  2π / ω

2 π √  m/ k  s  ∝  √ m

Hence  5/ 3   = √  m + 1 / m

m = 9 / 16 Kg

Thus the mass of the m is m = 9 / 16 Kg

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Answer 2

The mass of the body is 9/16 kg

Explanation:

From question, we know that, a body of mass m is attached to the lower in of a spring whose upper end is fixed the spring has negligible mass.

Now, the force for SHM of a hanging mass by a spring is given by the formula:

$\mathrm{F}=-k_{s x}$

Where,

$k_{s x}$ = μω²

⇒ $\omega=\sqrt{\frac{k_{s}}{m}}$

The time period of SHM is given by the formula:

T = 2π/ω

On substituting ω, we get,

T = 2π/$\sqrt{\frac{k_{s}}{m}}$ ∝ √m

Thus,

5/3 = √((m + 1)/m)

25/9 = (m + 1)/m

25 m = 9 m + 9

16 m = 9

∴ m = 9/16 kg