This set of Engineering Physics Multiple Choice Questions & Answers focuses on “Oscillations – 2”.
1. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be ___________
a) (πa√3)/T
b) (πa√3)/2T
c) πa/T
d) (3π2 a)/T
View Answer
Explanation: Speed,
v=ω√(a2-x2), x=a/2
v=ω√(a2-a2/4)=2π/T×(√3 a)/2=(πa√3)/T.
2. A body is executing simple harmonic motion. When the displacements from the mean position are 4cm and 5cm, the corresponding velocities of the body are 10cm/sec and 8cm/sec. Thenwhat is the time period of the body?
a) 2π sec
b) π/2 sec
c) π sec
d) 3π/2 sec
View Answer
Explanation: T=2π√(((x2)2-(x1)2)/((u1)2-(u2)2))=2π√((52-42)/(102-82))
T=2π√(9/36)=2π×3/6=πsec.
3. The total energy of a particle performing simple harmonic motion depends on ___________
a) k, a, m
b) k, a
c) k, a, x
d) k, x
View Answer
Explanation: E=1/2×mω2 a2=1/2×ka2
Clearly, E depends on k and a.
4. The particle executing simple harmonic motion has a kinetic energy K0 cos2 ωt. The maximum values of the potential energy and the total energy are respectively ___________
a) Kc/2 and K0
b) K0 and K0
c) K0 and 2K0
d) 0 and 2K0
View Answer
Explanation: When kinetic energy is maximum, potential energy is zero and vice-versa. But
Kinetic energy+ Potential energy = Total energy
Maximum potential energy = Maximum kinetic energy = Total energy = K0.
5. A linear harmonic oscillator of force constant 2×106 N/m and amplitude 0.01m has a total mechanical energy of 160J. Its ___________
a) Potential energy is 160 J
b) Potential energy is zero
c) Potential energy is 100J
d) Potential energy is 120J
View Answer
Explanation: Potential energy = 1/2×kx2
=1/2×2×106×0.012=100J.
6. The potential energy of a simple harmonic oscillation when the article is halfway to its end point is ___________
a) 2/3 E
b) 1/8 E
c) 1/4 E
d) 1/2 E
View Answer
Explanation: Total energy E = 1/2×ka2
At x = a/2, the potential energy is
Potential energy=1/2×k×(a/2)2
=1/4×1/2×ka2=E/4.
7. In a simple harmonic motion, when the displacement is one half the amplitude, what fraction of the total energy is kinetic?
a) 1/2
b) 3/4
c) Zero
d) 1/4
View Answer
Explanation: At x = a/2, the kinetic energy is
Kinetic energy=1/2×k×[a2-(a/2)2]=3/4×1/2×ka2=3/4 E.
8. A body executes simple harmonic motion with amplitude A. At what displacement from the mean position is the potential energy of the body one fourth of its total energy?
a) A/4
b) A/2
c) 3A/4
d) Some other fraction of A
View Answer
Explanation: Potential energy = 1/4×Total energy
1/2×kx2=1/4×1/2×kA2
x=A/2.
9. A loaded vertical spring executes simple harmonic motion with a time period of 4 sec. The difference between the kinetic energy and potential energy of this system varies with a period of?
a) 2 sec
b) 1 sec
c) 8 sec
d) 4 sec
View Answer
Explanation: Here T=4s. In one oscillation, both kinetic energy and potential energy become twice maximum and twice minimum. Hence the difference between kinetic energy and potential 2s.
10. A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4m is suspended from the same spring?
a) n/2
b) 4n
c) n/4
d) 2n
View Answer
Explanation: Here
n=1/2π×√(k/m)
n‘=1/2π×√(k/4m)=n/2.
Sanfoundry Global Education & Learning Series – Engineering Physics.
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