Engineering Physics Multiple Choice Questions 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?
a) 50%
b) 21%
c) 30%
d) 10.5%
Answer: d
Clarification: Time period,
T=2π√(l/g)
The percentage increase in a time period is given by,
∆T/T×100=1/2×∆l/l×100
∆T/T×100=1/2×21%=10.5%.
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 5cm is?
a) 6.25Nm
b) 12.50Nm
c) 18.75Nm
d) 25Nm
Answer: c
Clarification: W=1/2×k((x2)2-(x1)2)
W=1/2×5×103×(0.102-0.052)
W=18.75J.
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)?
a) E/4
b) E/16
c) E/2
d) E/8
Answer: a
Clarification: 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
b) f/2
c) f
d) 2f
Answer: a
Clarification: f=1/2π×√(k/m)
When spring is cut into two halves, spring constant of each half is 2k.
f‘=1/2π×√(2k/m)=√2 f.
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?
a) T/12
b) T/8
c) T/4
d) T/2
Answer: a
Clarification: y=asinωt
a/2=asin(2π/T)t
sin(2π/T)t=sin(π/6)
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. What is the amplitude of oscillations?
a) 0.28m
b) 0.36m
c) 0.707m
d) Zero
Answer: d
Clarification: amax=ω2 A
A=amax/ω2 = 7.5/(3.5×3.5)=0.61m.
7. The time period of a simple pendulum on a satellite, orbiting around the earth, is ___________
a) Infinite
b) Zero
c) 84.6 min
d) 24 hours
Answer: a
Clarification: In a satellite, g=0
T=2π√(l/g)=2π√(l/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?
a) T
b) T/10
c) √(9/10) T
d) √(10/9) T
Answer: d
Clarification: 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
T=2π√(l/(1-(ρ0/ρ))g)=2π√(l/(9/10)g)
T=√(10/9) T.
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?
a) k1/k2
b) √(k1/k2)
c) k2/k1
d) √(k2/k1)
Answer: d
Clarification: vmax (A)=vmax (B)
ω1 A1=ω2 A2
√(k1/m)×A1=√(k2/m)×A2
A1/A2 = √(k2/k1).
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 ___________
a) ω1=ω2
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