My Channel on YouTube → SimplePhysics with KK
For links to
other chapters - See bottom of the page
Or click here → kktutor.blogspot.com
SIMPLE HARMONIC MOTION:--
QUESTIONS FOR SHORT ANSWER
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.
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.
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.
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.
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.
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.
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.
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.
The figure for the Problem no-8
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.
9. Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?
ANSWER: The potential energy of an object is always with some zero potential energy reference point. If the mean position is chosen as a reference point then the potential energy will never be negative. It may be negative if some other point is selected as zero potential energy reference.
10. The energy of a system in simple harmonic motion is given by E=½m⍵²A². Which of the two statements is more appropriate?
(A) The energy is increased because the amplitude is increased.
(B) The amplitude is increased because the energy is increased.
ANSWER: The formula is for assessing the energy of a particle of mass m in SHM having amplitude A. If the mass is same the particle having larger amplitude will have greater energy. Though both statements are true the amplitude cannot increase itself until some work is done on the particle or some energy is given to it. So, the statement (B) is more appropriate.
11. A pendulum clock gives correct time at the equator. Will it gain time or lose time as it is taken to the poles?
ANSWER: At the poles, the value of g is more than the equator. The time period of the pendulum will decrease since T=2π√(l/g). It means the oscillations will be faster and hence the clock will gain time.
12. Can a pendulum clock be used in an earth satellite?
ANSWER: In an earth satellite the net acceleration due to the gravity is zero so the bob of the pendulum will be weightless and it will not oscillate. Thus the pendulum clock cannot be used in an earth satellite.
13. A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for the simple pendulum is valid with the distance between the point of suspension and center of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?
ANSWER: The time period is given as T=2π√(l/g).
It is clear from this expression that the time period depends only upon the effective length of the pendulum as g is constant at a place.
The figure for Problem number - 13
Initially, the effective length of the pendulum l is the distance between the center of the bob to the point of support. As the water leaks out the center of mass of the water in the bob goes down. So the combined center of mass of the bob also goes down, thus the effective length increases. Thus the time period begins to increase. Though the CoM of the water goes down till the last drop the mass of the water continuously goes reducing till zero. So the combined center of mass after going down to a certain point again begins to go up till it reaches the center of the hollow sphere. Thus the effective length of the pendulum after reaching a maximum slowly reduces back to its original length. The effect is that the time period of the pendulum slowly increases and then reaching a maximum begins to reduce and finally its value becomes the original value.
14. A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?
ANSWER: Yes.
The time period of the vertical oscillation is given by
T =2π√(m/k)
When the block is in equilibrium,
mg=ks {where s is the extension of the spring}
→k=mg/s
Thus T=2π√(ms/mg) =2π√(s/g)
Thus the time period of the vertical oscillation can be found out by measuring the extension of the spring when the block is in equilibrium.
15. A platoon of soldiers marches on a road in steps according to the sound of a Marching band. The band is stopped and the soldiers are ordered to break the steps while crossing a bridge. Why?
ANSWER: The march in step is a type of forced oscillation of the bridge. If the frequency of the steps be near to the frequency of the bridge the resonance may occur that will result in a very large amplitude of the bridge oscillation. It will be damaging to the bridge. Hence the march on the bridge is by breaking the steps.
16. The force acting on a particle moving along X-axis is F = -k(x-v₀t) where k is a positive constant. An observer moving at a constant velocity v₀ along the X-axis looks at the particle. What kind of motion does he find for the particle?
ANSWER: F= -kx+kv₀t
The force has two part. One is dependent on displacement and the other is dependent on the time. (The first part is negative while the other is positive, the second part is ever-increasing.) So it is not a simple harmonic motion. The observer is moving with a constant velocity so he still observes from an inertial frame. The observer will also find that it is not a simple harmonic motion.
===<<<O>>>===
Links to the chapters -
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
 |
| The figure for the Problem no-8 |
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.
9. Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?
ANSWER: The potential energy of an object is always with some zero potential energy reference point. If the mean position is chosen as a reference point then the potential energy will never be negative. It may be negative if some other point is selected as zero potential energy reference.
10. The energy of a system in simple harmonic motion is given by E=½m⍵²A². Which of the two statements is more appropriate?
(A) The energy is increased because the amplitude is increased.
(B) The amplitude is increased because the energy is increased.
(B) The amplitude is increased because the energy is increased.
ANSWER: The formula is for assessing the energy of a particle of mass m in SHM having amplitude A. If the mass is same the particle having larger amplitude will have greater energy. Though both statements are true the amplitude cannot increase itself until some work is done on the particle or some energy is given to it. So, the statement (B) is more appropriate.
11. A pendulum clock gives correct time at the equator. Will it gain time or lose time as it is taken to the poles?
ANSWER: At the poles, the value of g is more than the equator. The time period of the pendulum will decrease since T=2π√(l/g). It means the oscillations will be faster and hence the clock will gain time.
12. Can a pendulum clock be used in an earth satellite?
ANSWER: In an earth satellite the net acceleration due to the gravity is zero so the bob of the pendulum will be weightless and it will not oscillate. Thus the pendulum clock cannot be used in an earth satellite.
13. A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for the simple pendulum is valid with the distance between the point of suspension and center of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?
ANSWER: The time period is given as T=2π√(l/g).
It is clear from this expression that the time period depends only upon the effective length of the pendulum as g is constant at a place.
Initially, the effective length of the pendulum l is the distance between the center of the bob to the point of support. As the water leaks out the center of mass of the water in the bob goes down. So the combined center of mass of the bob also goes down, thus the effective length increases. Thus the time period begins to increase. Though the CoM of the water goes down till the last drop the mass of the water continuously goes reducing till zero. So the combined center of mass after going down to a certain point again begins to go up till it reaches the center of the hollow sphere. Thus the effective length of the pendulum after reaching a maximum slowly reduces back to its original length. The effect is that the time period of the pendulum slowly increases and then reaching a maximum begins to reduce and finally its value becomes the original value.
It is clear from this expression that the time period depends only upon the effective length of the pendulum as g is constant at a place.
 |
| The figure for Problem number - 13 |
Initially, the effective length of the pendulum l is the distance between the center of the bob to the point of support. As the water leaks out the center of mass of the water in the bob goes down. So the combined center of mass of the bob also goes down, thus the effective length increases. Thus the time period begins to increase. Though the CoM of the water goes down till the last drop the mass of the water continuously goes reducing till zero. So the combined center of mass after going down to a certain point again begins to go up till it reaches the center of the hollow sphere. Thus the effective length of the pendulum after reaching a maximum slowly reduces back to its original length. The effect is that the time period of the pendulum slowly increases and then reaching a maximum begins to reduce and finally its value becomes the original value.
14. A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?
ANSWER: Yes.
The time period of the vertical oscillation is given by
T =2π√(m/k)
When the block is in equilibrium,
mg=ks {where s is the extension of the spring}
→k=mg/s
Thus T=2π√(ms/mg) =2π√(s/g)
Thus the time period of the vertical oscillation can be found out by measuring the extension of the spring when the block is in equilibrium.
The time period of the vertical oscillation is given by
T =2π√(m/k)
When the block is in equilibrium,
mg=ks {where s is the extension of the spring}
→k=mg/s
Thus T=2π√(ms/mg) =2π√(s/g)
Thus the time period of the vertical oscillation can be found out by measuring the extension of the spring when the block is in equilibrium.
15. A platoon of soldiers marches on a road in steps according to the sound of a Marching band. The band is stopped and the soldiers are ordered to break the steps while crossing a bridge. Why?
ANSWER: The march in step is a type of forced oscillation of the bridge. If the frequency of the steps be near to the frequency of the bridge the resonance may occur that will result in a very large amplitude of the bridge oscillation. It will be damaging to the bridge. Hence the march on the bridge is by breaking the steps.
16. The force acting on a particle moving along X-axis is F = -k(x-v₀t) where k is a positive constant. An observer moving at a constant velocity v₀ along the X-axis looks at the particle. What kind of motion does he find for the particle?
ANSWER: F= -kx+kv₀t
The force has two part. One is dependent on displacement and the other is dependent on the time. (The first part is negative while the other is positive, the second part is ever-increasing.) So it is not a simple harmonic motion. The observer is moving with a constant velocity so he still observes from an inertial frame. The observer will also find that it is not a simple harmonic motion.
===<<<O>>>===
Links to the chapters -
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
HC Verma's Concepts of Physics, Chapter-7, Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
HC Verma's Concepts of Physics, Chapter-6, Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
HC Verma's Concepts of Physics, Chapter-4, The Forces
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
No comments:
Post a Comment