Q#1
A person goes to bed at sharp 10.00 pm every day. Is it an example of periodic motion? If yes, what is the time period? If no, why?Answer:
Going to the bed every day at 10.00 pm is an event, not a motion.
Q#2
A particle executing Simple Harmonic Motion comes to rest at the extreme positions. Is the resultant force on the particle zero at these positions according to Newton's First Law?
Answer:
In a simple harmonic motion the magnitude of the resultant force is proportional to the displacement from the mean position. At the extreme positions, the displacement is maximum hence the magnitude of the force is also maximum.
From Newton's First Law we infer that if a body remains at rest then the resultant force on it is zero but in this case, the body comes to rest momentarily and moves back towards the mean position because the force and acceleration are not zero at extreme points.
Q#3
Can simple harmonic motion take place in a noninertial frame? If yes should the ratio of the force applied with the displacement be constant?
Answer:
Yes. Consider an accelerating vehicle in which a block is attached to a spring on a smooth horizontal surface and the other end is fixed. The rest position of the block will change compared to the nonaccelerating vehicle. If the block is stretched and released it will be in a simple harmonic motion having the mean position changed to a new one.
But the ratio of the applied force with the displacement will not be constant until we take a pseudo force into account that is mass times the acceleration of the vehicle and opposite to the direction of the accelerating frame.
Q#4
A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is the displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?
Answer:
No. Only we can say that the particle is at the extreme point if the velocity at that instant is zero. If the velocity is maximum we can say that it is at the mean point and the displacement from the mean point is zero.
Q#5
A small creature moves with constant speed in a vertical circle on a bright day. Does its shadow formed by the sun on a horizontal plane move in a simple harmonic motion?
Answer:
Yes. The projection of a point in a uniform circular motion on a diameter is a simple harmonic motion.
Q#6
A particle executes simple harmonic motion. Let P be a point near the mean position and Q be a point near an extreme. The speed of the particle at P is larger than the speed at Q. Still the particle crosses P and Q equal number of times in a given time interval. Does it make you unhappy?
Answer:
No. Since the particle at P and Q is not in uniform motion but in SHM and the same particle is in different phases at these points. Hence it is quite ok.
Q#7
In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.
Answer:
The speed near the extreme position is near zero and the decision by the observer that the bob has reached the extreme position may have some error due to slow movement. But the bob crosses the mean position with maximum speed so there is no hesitation in deciding that the bob has crossed the mean position and hence better accuracy.
Q#8
It is proposed to move a particle in simple harmonic motion on a rough horizontal surface by applying an external force along the line of motion. Sketch the graph of the applied force against the position of the particle. Note that the applied force has two values for a given position depending on whether the particle is moving in a positive or negative direction.
Answer:
Let us assume that the coefficient of kinetic friction µ is constant throughout the movement. If the simple harmonic motion is between points A and B with O as a mean point then amplitude OA = OB = a. See the figure below.
When the particle is crossing O from left to right, the total force on it is equal to F-µmg. It should be zero. Thus, F = µmg (OC in the figure). In between O and A, F - µmg = -m⍵²x. The equation of a straight line. Just before stopping at A, F = µmg - m⍵²a (Represented as AD). When it stops at A instantaneously, µmg = 0, thus F = -m⍵²a. But when it begins to move back from A,
F + µmg = -m⍵²a
F = -µmg - m⍵²a
(represented by AD')
So between just approaching A and departing A the value of the magnitude of the force changes from µmg-m⍵²a towards the right to µmg+m⍵²a towards the left.
When crossing O from right to left
F + µmg = 0
F = -µmg (Represented by OC').
Similarly, the rest of the graph can be plotted as in the figure.
F + µmg = -m⍵²a
F = -µmg - m⍵²a
(represented by AD')
So between just approaching A and departing A the value of the magnitude of the force changes from µmg-m⍵²a towards the right to µmg+m⍵²a towards the left.
When crossing O from right to left
F + µmg = 0
F = -µmg (Represented by OC').
Similarly, the rest of the graph can be plotted as in the figure.
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