OSCILLATIONS AND WAVES

6 ) Free Physics Questions Online Mock Test Series By Cracktest

As shown in figure, two masses of 3.0 kg and 1.0 kg are attached at the two ends of a spring having force constant 300 Nm^–1. The natural frequency of oscillation for the system will be ………….. Hz. (Ignore friction)

1/4

1/3

View Solution

Answer :

Reduced mass of the system = m = {(m₁ ∙ m₂) / (m₁ + m₂)}

= {(3 × 1) / (3 + 4)} = (3/4) kg.

as T = 2π √(m/k) hence natural frequency f = (1/2π) √(k/m)

∴ f = (1/2π) √{(300) / (3/4)}

= (1/2π) √(400)

= (20 / 2π)

= (10 / π) = 3.18 Hz

≈ 3Hz

7 )

If (1/4) of a spring having length ℓ is cutoff, then what will be the spring constant of remaining part?

4k / 3

3k / 4

View Solution

Answer :

k₁ℓ₁ = k₂ℓ₂ = kℓ

where ℓ is complete length

ℓ₁ = (ℓ/4)

ℓ₂ = (3ℓ / 4) (remaining part)

∴ K₁ (ℓ/4) = k₂ (3ℓ / 4) = kℓ

∴ k₂ = (4k / 3)

8 )

The amplitude for a S.H.M. given by the equation x = 3 sin 3 pt + 4 cos 3 pt is ……….… m.

View Solution

Answer :

amplitude of SHM is given for x = a sin ωt + b cos ωt as

A = √(a² + b²)

here x = 3 sin 3 pt + 4 cos 3 pt

∴ A = √(3² + 4²) = 5

9 )

When an elastic spring is given a displacement of 10mm, it gains an potential energy equal to U. If this spring is given an additional displacement of 10 mm, then its potential energy will be………….

U/4

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Answer :

PE = potential energy ∝ (displacement)²

when PE = U, displacement = 10 mm

hence when displacement = 10 + 10 = 20mm, then potential energy be U'

hence

(U' / U) = (20 / 10)²

U' = (2)² U

U' = 4U

10 )

The increase in periodic time of a simple pendulum executing S.H.M. is…………….. when its length is increased by 21%.

10%

11%

12%

13%

View Solution

Answer :

T = 2π √(ℓ/g) i.e. T ∝ √ℓ

∴ (T₂ / T₁) = √(ℓ₂ / ℓ₁)

∴ T₂ = T₁ √(ℓ₂ / ℓ₁)

here ℓ₂ = ℓ₁ (1.21)

∴ T₂ = T₁ √{(1.21 ℓ₁) / ℓ₁}

∴ T₂ = T₁ √(1.21)

T₂ = 1.1 T₁

i.e. {(T₂ – T₁) / T₁} = [{1.1 T₁ – T₁} / T₁] = 0.1

∴ % increase in periodic time = 10%

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