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A wave is essentially an oscillation that travels (picture a pendulum that always pays attention in class, asks relevant questions and gets straight A’s).
The recurring conversion of potential energy to kinetic energy and back to potential energy distinguishes SHM. Consider the following to go back to our pendulum: As you bring the pendulum back, its gravitational potential energy grows. When you let go, the gravitational potential energy is transformed to kinetic energy. The kinetic energy is largest near the bottom of the pendulum’s oscillation; when the pendulum moves upward, the kinetic energy is converted back to gravitational potential energy.
Let’s keep thinking about energy as we switch our attention to waves. Consider throwing a stone into a pond to generate energy. The stone will disturb the pond’s surface and the waves in the water will convey energy away from the pebble that is causing the disturbance. Consider the case of a single water molecule that landed a little distance from the stone. The wave will push this water molecule higher (gaining potential energy), then lower (gaining potential energy); it will pass a point of maximal kinetic energy, then gain potential energy when it is pushed below the level of the resting surface.
Hopefully, when the wave passes, it will become clear that this molecule is undergoing SHM! Unlike the wave, however, the water does not move away from the impact point; rather, the medium oscillates in position as the wave passes through.
1. In SHM, choose the right statement for a particle’s force.
a) It is proportional to velocity in a linear manner.
b) It is proportional to position in a linear way.
c) It is angled away from the centre of gravity.
d) It’s moving in the right direction.
Answer: b) It is proportional to position in a linear way.
A particle’s force is always directed toward its mean position. When x = Asin(ωt),
F = -mAω2sin(ωt). As a result, force is proportional to position in a linear fashion.
2. F= -kx is the force on a particle of mass ‘m’ undergoing SHM. What is the relationship between x and in terms of angular frequency?
a) k = ω2 ω
b) k = m√ω
c) m = k/ω2
d) m = k2/ω
Answer: c) m = k/ω2
In SHM, the force on a particle equals -mw2x (t).
In this scenario: F = -kx (t).
k=mω2 is the result. Alternatively, m = k/ω2
3. The force applied on a 1kg particle is -2x, where x is the displacement from SHM’s mean location. What will be the time period of the oscillations?
a) 2 πs
b) π√2 s
c) π s
d) 2√2π s
Answer: b) π√2 s
We get 2 = 1ω2 or w = √2 when we use k=mω2. As a result, the time period = 2π/ω=π√2 s.
4. A particle is now executing SHM and moving towards the amplitude. What is the relationship between the direction of velocity and acceleration if it is at A/2?
a) Both vectors are pointing in the direction of the amplitude.
b) The direction of velocity is towards the amplitude, whereas the direction of acceleration is towards the mean position.
c) Acceleration is directed towards the amplitude, while velocity is directed towards the mean position.
d) Both vectors are pointing in the direction of the mean position.
Answer: b) The direction of velocity is towards the amplitude, whereas the direction of acceleration is towards the mean position.
In SHM, a particle’s force is always directed towards the mean position. Because the particle is currently approaching an extreme, its velocity will be in the direction of the amplitude.
5. Will the magnitude of maximum acceleration in a SHM be greater than the magnitude of maximum velocity for what value of w? The angular frequency is denoted by w.
a) ω > 1
b) ω < 1
c) ω = 0
d) Not possible for any value of ω
Answer: a) ω > 1
Let’s say SHM’s velocity equation is v = Aωcos(ωt), and the acceleration equation is
a = -Aω2sin (ωt). A is the maximum magnitude of vel. Aω2 is the maximum magnitude of acc. For example, Aω2>Aω , ω>1.
6. At t=0, a particle of mass m begins at the mean position of a SHM and moves towards -A. Find the force acting on SHM as a function of time if its angular frequency is w.
a) mAw2sin(wt)
b) mAw2sin(wt+)
c) -mAw2cos(wt+)
d) -mAw2cos(wt)
Answer: a) mAw2sin(wt)
Displacement can be given by: x = -Asin (wt).
We get v= -Awcos(wt) when we take the derivative of displacement.
We also get: a = Aw2sin (wt).
Hence, force will be: F(t)=mAw2sin(wt)
7. A particle is undergoing a SHM with a 10cm amplitude. What should the minimum acceleration be at an extreme point for the maximum speed at the centre to be 5 metres per second?
a) 20m/s2
b) 5m/s2
c) 0
d) 250m/s2
Answer: d) 250m/s2
The value of vmax will be vmax =Aw
5 = 0.1w.
w = 50 s-1
amin at extreme will be = Aw2
= 0.1×50×50
=250m/s2
8. The diameter of a particle in uniform circular motion is projected. The projection will move in a simple harmonic motion. Choose the proper choice for the particle’s speed and acceleration in circular motion.
a) The speed is constant and the acceleration is nothing.
b) The speed remains constant, but the acceleration is not zero.
c) Acceleration is non-zero and speed changes.
d) The speed changes, but there is no acceleration.
Answer: a) The speed is constant and the acceleration is nothing.
The magnitude of velocity remains constant for a particle moving in a uniform circular motion, but the direction is constantly changing. As a result, while speed is constant, acceleration is not.
9. What is an oscillator’s proportionality constant if the damping force is proportional to the velocity?
a) kg.s-1
b) kg.m.s-1
c) kg.s
d) kg.m.s-2
Ans a) kg.s-1
10. What is the phase difference between the tuning fork’s prongs?
a) 5π
b)π
c) 2π
d) 3π
Answer: b)π
11. When a wave is reflected from a denser medium, how does the phase change?
a) 3π
b) 0
c) 2π
d) π
Answer: d) π
12. The observer perceives a sound coming from a source that is moving away from the stationary observer. Determine the sound’s frequency.
a) It’s been halved.
b) It hasn’t changed.
c) It extends to infinity.
d) It’ll be doubled.
Answer: a) It’s been halved.
