This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Oscillatory Motion – 2”.
1. The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increases length is
Explanation: Time period,
The percentage increase in a time period is given by,
2. A spring of spring constant 5×103N/m is stretched initially by 5cm from the unstretched position. Then the work done to stretch is further by another 5cmis
3. A particle is executing simple harmonic motion at midpoint of mean position and extremely. What is the potential energy in terms of total energy (E)?
Explanation: At y = A/2,
Potential energy = 1/2×kx2
Potential energy=1/2×k×A2/22 =1/4×1/2×k×A2=1/4×E.
4. A mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into two halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation of mass will be
a) √2 f
When spring is cut into two halves, spring constant of each half is 2k.
5. A particle executes simple harmonic motion with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
2π/T×t=π/6 or t=T/12.
6. A particle executes simple harmonic motion with an angular velocity of 3.5 rad/sec and maximum velocity acceleration 7.5 m/s2 respectively. The amplitude of oscillations is
Explanation: amax=ω2 A
7. The time period of a simple pendulum on a satellite, orbiting around the earth, is
c) 84.6 min
d) 24 hours
Explanation: In a satellite, g=0
8. A simple pendulum has a time period T. The pendulum is completely immersed in a non-viscous liquid, whose density is 1/10th of that of the material of the bob. The time period of the pendulum is immersed in the liquid is
c) √(9/10) T
d) √(10/9) T
Explanation: In air, T=2π√(l/g)
Let ρ be the density of the bob material. When the bob is immersed in a non-viscous liquid of density ρ0=ρ/10, time period becomes
9. Two bodies M and N of equal masses are suspended from separate massless spring of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of variation of M to that of N is
d) √(k2/k1 )
Explanation: vmax (A)=vmax (B)
ω1 A1=ω2 A2
10. In forced oscillation of a particle, the amplitude is maximum for a frequency ω1 of the force, while the energy is maximum for a frequency ω2 of the force. Then
b) ω1 is lessed than ω2
c) ω1 is lesser than ω2, when damping is small and ω1 is greater than ω2, when damping is large
d) ω1 is lesser than ω2
Explanation: Only in case of resonance, both the amplitude and energy of oscillation are maximum. Hence, ω1=ω2.
11. If a simple pendulum oscillates with amplitude of 50mm and time period of 2s, then its maximum velocity is
12. If the period of oscillation of mass m suspended from a spring is 2s, then the period of mass 4m will be
Explanation: T2/T1 =√(m2/m1)=√(4m/m)=2
13. Statement: Resonance is a special case of forced vibration in which the nature and frequency of vibration of the body is the same as the impressed frequency and the amplitude of forced vibration, is maximum
Reason: The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force
a) Both statement and reason are true and the reason is the correct explanation of the statement
b) Both statement and reason are true but the reason is not the correct explanation of the statement
c) Statement is true, but the reason is false
d) Statement and reason are false
Explanation: The statement is true but the reason is false. The amplitude of forced vibrations increases when the frequency of the impressed force approaches the natural frequency of the driven body.
Sanfoundry Global Education & Learning Series – Engineering Physics.
To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.