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Practice each question, then open the answer or worked solution.
What are the two basic characteristics of a simple harmonic motion?
The following are the two basic characteristics of an SHM:
a) acceleration and displacement are directly proportional to each other
b) the direction of acceleration is towards the mean position, while the direction of displacement is towards the opposite side of the mean position
When will the motion of a simple pendulum be simple harmonic?
Consider a simple pendulum with O as the fixed point that has a restoring torque.
The restoring force is given as F = -mg sin θ
θ is so small that θ = arc/radius = x/l
Therefore, F = – mg x/l
Therefore, it can be said that the motion of the simple pendulum will be simple harmonic when θ is small.
What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?
The displacement equation of SHM is x = a sin (ꞷt + ϕ)
The velocity of the particle, v = dx/dt = d a sin (ꞷt + ϕ)/dt
Maximum velocity, |v|max = aꞷ
Acceleration, A = dv/dt = aꞷ2 sin (ꞷt + ϕ)
Maximum acceleration, |A|max = ꞷ2a
|v|max/|A|max = ꞷ
What is the ratio between the distance travelled by the oscillator in one time period and amplitude?
Distance travelled by the oscillator in one time period = 4A
Where A is the amplitude of the oscillation
Therefore, required ration = 4A/A = 4/1 = 4:1
In the figure, what will be the sign of the velocity of the point P’, which is the projection of the velocity of the reference particle P. P is moving in a circle of radius R in the anticlockwise direction?
P’ is the perpendicular of the velocity vector for the particle Pat time t.
When the particle moves from P to P1, the foot of the particle moves from P’ to Q, and that is towards the negative axis.
Therefore, the sign is negative.
Show that for a particle executing SHM, velocity, and displacement have a phase difference of π/2.
The displacement equation of SHM, x = a cos ꞷt
Velocity, v = dx/dt = -aꞷ sin ꞷt
The phase displacement, ϕ1 = ꞷt
Phase velocity, ϕ2 = π/2 + ꞷt
Therefore, difference in phase of displacement = ϕ2 – ϕ1 = π/2
Draw a graph to show the variation of PE, KE, and total energy of a simple harmonic oscillator with displacement.
The length of a second’s pendulum on the surface of the earth is 1 m. What will be the length of a second’s pendulum on the moon?
The time period of a simple pendulum, T = 2π√l/g
Where l is the length of the pendulum and g is the acceleration due to gravity
Therefore, lmoon/gmoon = learth/gearth = 1/6 m
Short AnswersFind the time period of mass M when displaced from its equilibrium position and then released for the system shown in the figure.
The distance through which the mass is pulled is y.
The total extension of the spring is y + y = 2y.
The initial equilibrium mean position = x0
The net extension in the spring = x0 + 2y
When the equation, F = 2T is solved where T = kx0,
We get, a as proportional to –x.
When the acceleration obtained above is compared to the standard SHM equation, we get
ꞷ2 = 4k/m
ꞷ = √4k/M
Show that the motion of a particle represented by y = sin ωt – cos ωt is simple harmonic with a period of 2π/ω.
A function will represent SHM and is written as sin (2π/T t + ϕ)
y = sin ωt – cos ωt
y = √2 sin (ωt – π/4)
The standard SHM has y = a s sin (2π/T t + ϕ)
T = 2π/ω