Engineering Physics online test focuses on “Wave Motion – 2”.
1. A particle in simple harmonic motion is described by the displacement function x(t)=Acos(ωt+θ). If the initial (t=0) position of the particle is 1cm and its initial velocity isπcm/s, what is its amplitude? The angular frequency is the particle is πrad/s.
a) 1 cm
b) √2 cm
c) 2 cm
d) 2.5 cm
Answer: b
Clarification: v=ω√(A2-x2)
π=π√(A2-1)
A2-1=1 or A2=2
A=√2cm.
2. A particle executes simple harmonic motion, its time period is 16s. If it passes through the centre of oscillation, then its velocity is 2 m/s at time 2s. The amplitude will be ___________
a) 7.2m
b) 4cm
c) 6cm
d) 0.72m
Answer: a
Clarification: Here,
t=2s,v=2m/s,T=16s
v=Aωcosωt
2=A×2π/14×cos(2π/16×2)
A=(16√2)/π=7.2m.
3. A body is executing the simple harmonic motion with an angular frequency of 2rad/sec. Velocity of the body at 20m displacement, when amplitude of motion is 60m, is ___________
a) 90 m/s
b) 118 m/s
c) 113 m/s
d) 131 m/s
Answer: c
Clarification: v=ω√(A2-y2)=2√(602-202)
v=80√2
v=113m/s.
4. A particle is executing simple harmonic motion of amplitude 10cm. Its time period of oscillation is π seconds. The velocity of the particle when it is 2 cm from extreme position is ___________
a) 10 cm/s
b) 12 cm/s
c) 16√16 cm/s
d) 16 cm/s
Answer: b
Clarification: v=2π/T×√(A2-y2)
v=2π/π √(102-82)
=2×6=12cm/s.
5. The magnitude of acceleration of particle executing simple harmonic motion at the position of maximum displacement is?
a) Zero
b) Minimum
c) Maximum
d) Infinity
Answer: c
Clarification: Acceleration in simple harmonic motion is
a=ω2 y
At ymax=A, amax=ω2 A
Acceleration is maximum at the position of maximum displacement.
6. The maximum velocity and maximum acceleration of a body moving in a simple harmonic motion are 2m/s and 4m/s2 respectively. What will be the angular velocity?
a) 4 rad/sec
b) 3 rad/sec
c) 2 rad/sec
d) 8 rad/sec
Answer: c
Clarification: vmax=ωA, amax=ω2 A
ω=amax/vmax = 4/2
ω=2rad/sec.
7. A particle executing simple harmonic motion has amplitude 0.01 and frequency 60Hz. The maximum acceleration of the particle is ___________
a) 144 π2 m/s2
b) 80 π2 m/s2
c) 120 π2 m/s2
d) 60 π2 m/s2
Answer: a
Clarification: amax=ω2 A=4π2 v2 A
=4π2×60×60×0.01=144 π2 m/s2.
8. A particle having potential energy 1/3 of the maximum value at a distance of 4 cm from mean position. Amplitude of motion is ___________
a) 4√3
b) 6/√2
c) 2/√6
d) 2√6
Answer: a
Clarification: Ep=1/3 E
1/2 ky2=1/3×1/2×kA2
A=√3 y=√3×4=4√3 cm.
9. A particle executes simple harmonic motion of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy?
a) 0.51A
b) 0.71A
c) 0.61A
d) 0.81A
Answer: b
Clarification: Ek=Ep
1/2 k(A2-y2)=1/2×ky2
y=±A/√2
y=±0.71A.
10. To show that a simple pendulum executes simple harmonic motion, it is necessary to assure that ___________
a) Length of the pendulum is small
b) Amplitude of oscillation is small
c) Mass of the pendulum is small
d) Acceleration due to gravity is small