Question

Problem - 3
A load of 100 g increases a length of light spring by 10 cm. Find the
period of its vertical oscillations when a mass of one kg is attached to
the free end of the spring. Take g = 10 m/s
​

Answer 1

Given :

• Mass of the load(m₁) = 100 g = 0.1 kg
• Increase in length of the spring(x) = 10 cm = 0.1 m
• Acceleration due to gravity (g) = 10 m/s ²

• New mass of the load (m₂) = 1 kg

To find :

• New time period

Solution :

The time period of oscillation of a spring is given by the formula ,

T = 2π√m/k

here ,

• T = time period
• m = mass
• k = spring constant .

Force acting on the load in down ward direction ,

• F = mg

The magnitude of the restoring force acting on the load by the spring ,

• F = kx

Hence ,

⇒ kx = mg

⇒ k = m x g / x

⇒ k = 0.1 x 10 / 0.1

⇒ k = 10

Hence ,

⇒ T₁ = 2π√m₁/k

⇒ T₁ = 2 x 3.14 √0.1 / 10

⇒ T₁ = 6.28 x √0.01

⇒ T₁ = 6.28 X 0.1

⇒ T₁ = 0.628 s

As 2 , π and k are constants we can say that ,

T ∝ √m

Hence ,

⇒ T₁ / T₂ = √m₁ / m₂

⇒  T₁ / T₂ = √0.1 / 1

⇒ T₁ / T₂ = 0.316

⇒ T₂ =  T₁ / 0.316

⇒ T₂ =  0.628 / 0.316

⇒ T₂ = 1.98  s

The new time period of oscillation is 1.98 s .