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SIMPLE HARMONIC MOTION:--
EXERCISES Q31 TO Q40
31. A uniform plate of mass M stays horizontally and symmetrically on two wheels rotating in opposite directions (Figure 12-E16). The separation between the wheels is L. The friction coefficient between each wheel and the plate is µ. Find the time period of oscillation of the plate if it is slightly displaced along its length and released.
Figure for Q-31
ANSWER: When the center of the plate is above the midpoint of the line joining the wheels, both wheels get equal weight and the friction force by both the wheels are equal. Let the slight displacement along the length be d. If the normal force by the wheels are R and R', then
R+R' =Mg
and R*L = Mg*(d+L/2) {Taking torque about the right wheel}
→R =½Mg(2d+L)/L
Figure for Q-31
And R' = Mg-R
=Mg - ½Mg(2d+L)/L
=½[2MgL-MgL-2Mgd]/L
=½[MgL-2Mgd]/L
So net friction force on the plate F = µ(R-R')
=(µ/2L)[2Mgd+MgL-MgL+2Mgd]
=(µ/2L)[4Mgd]
=2µMgd/L
→F = M(2µg/L)d
Which is in the form of F =m⍵²r, ie. the force is proportional to the displacement. So it is a simple harmonic motion with
⍵ = √(2µg/L)
Hence the time period T = 2π/⍵ =2π√(L/2µg)
32. A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where g = π² m/s².
ANSWER: Time period T =2 s. But the time period of a simple pendulum is given by T = 2π√(l/g).
→l/g = T²/4π²
→l = gT²/4π²
putting the values,
l = π²*4/4π² = 1 m
33. The angle made by the string of a simple pendulum with the vertical depends on time as θ = (π/90).sin[(π s⁻¹)t]. Find the length of the pendulum if g = π² m/s².
ANSWER: The given equation θ = (π/90).sin[(π s⁻¹)t]
is in the form of the simple harmonic motion displacement A.sin⍵t. From comparison,
⍵ = π s⁻¹.
Time period T = 2π/⍵ =2π/π = 2 s
But for the simple pendulum T =2π√(l/g)
→2 = 2π√(l/π²) =2√l
→l = 1 m
34. The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours?
ANSWER: The pendulum of a clock has a correct time period of 2 s but this clock takes more time in one oscillation. So it is running slow. Each second it is slower by 2.04/2 -1 =0.02 s. Hence in 24 hr i.e. 24*3600 s it runs 24*3600*0.02 s =0.48*3600/3600 hr =0.48*60 minutes =28.80 minutes slow.
35. A pendulum clock giving correct time at a place where g = 9.800 m/s² is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.
ANSWER: The pendulum loses 24 s in 24 hrs, i.e. 1 s each hr or 3600 s. The correct clock has a time period T, of 2 s which makes 3600/2 =1800 oscillations per hour. But this clock is taking 3600+1 =3601 s in 1800 oscillations (In one hour). So the new time period T' = 3601/1800 s
If the value of g at the new place = g'
then, T' = 2π√(L/g') and T =2π√(L/g), dividing
T'/T = √(g/g')
→g' =gT²/T'² =9.80*2²/(3601/1800)²
=9.80*4/(2.0005)²
=9.795m/s²
36. A simple pendulum is constructed by hanging a heavy ball by 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system the system is taken on the moon where the acceleration due to the gravitation of the moon is 1.67 m/s²?
ANSWER: (a) l = 5.0 m, So T =2π√(l/g) =2π√(5/9.8) =1.429π seconds.
Oscillations per seconds = frequency =1/T = 1/1.429π Hz
=0.70/π Hz or oscillations per second.
(b) At moon g =1.67 m/s²
The frequency = 1/T= 1/{2π√(l/g)}
=1/{2π√(5/1.67)}
=1/(2π√3) Hz
37. The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.
ANSWER: The maximum tension will be when the string is vertical where it balances the weight as well as the centrifugal force. If the speed of the bob at this position is v, Tₘₐₓ = mg+mv²/l
The minimum tension in the string will be at the highest point of the oscillation where the speed of the bob is zero. Tₘᵢₙ =mg.cosθ.
Figure for Q-37
Given that Tₘₐₓ = 2*Tₘᵢₙ
→mg+mv²/l = 2mg.cosθ
→cosθ = ½(1+v²/gl) .... (i)
Since in a SHM the total energy is constant at any instant, equating the total energies at these two positions we have,
½mv² = mg(l-l.cosθ) .........{Taking reference for P.E. at the lowest point of the bob}
→v² = 2gl(1-cosθ)
Putting in (i)
cosθ = ½{1+2(1-cosθ)}=½(3-2.cosθ) =3/2 - cosθ
→2.cosθ =3/2
→cosθ =3/4
→θ =cos⁻¹(3/4)
38. A small block oscillates back and forth on a smooth concave surface of radius R (figure 12-E17). Find the time period of small oscillation.
The figure for Q - 38
ANSWER: Let the mass of the mall block = m,
The normal force on the block = mg.cosθ where θ is the angle between the radius at any instant t with the vertical. The tangential force F = mg.sinθ
Acceleration of the block = a = mg.sinθ/m = g.sinθ
If the angular displacement θ is small then sinθ= θ and the block moves in approximately a straight line.
Now acceleration a =g.sinθ =gθ
If the displacement of the block from the lowest position is x, then θ=x/R
So, a =gx/R =(g/R)x
g/R is constant, hence the acceleration is proportional to the displacement. Thus it is a simple harmonic motion with ⍵² =g/R
→⍵ = √(g/R)
The time period of the oscillation =2π/⍵ =2π√(R/g)
39. A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillation about the lowest point. Find the time period.
ANSWER: Let the ball be displaced a distance x from the mean position. If the radius of the surface at this place makes an angle θ with the vertical, the restoring force on the ball = mg.sinθ
Since the ball rolls on the surface without slipping the torque of this force about the point of contact = mgr.sinθ
Moment of inertia of the ball about an axis passing through its center = 2mr²/5
M.I. of the ball about the axis passing through the point of contact
I = 2mr²/5 +mr² =7mr²/5
Angular acceleration α = Torque/ M.I.
=mgr.sinθ/(7mr²/5)
=5g.sinθ/7r
Linear acceleration of the center of ball, a=αr =5g.sinθ/7 ---------(ii)
If the angle θ is small, sinθ =θ =x/(R-r)
so, a = 5gx/7(R-r) ={5g/7(R-r)}x
Hence the acceleration is proportional to the displacement and it is in simple harmonic motion with ⍵² ={5g/7(R-r)}
Hence the time period T =2π/⍵ =2π√{7(R-r)/5g}
40. A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep, calculate the time period of the pendulum there. Radius of the earth = 6400 km.
ANSWER: R = 6400 km
h = 1600 km, l=0.40 m
The gravitational field at the depth of 1600 km =GM'/(R-h)²
M' is the mass of the earth part with radius R-h.
The Force on the bob of the pendulum if its mass is m
= GM'm/(R-h)²
If the acceleration due to the gravity there =g'
Then mg' =GM'm/(R-h)²
g' = GM'/(R-h)²
If the density of the earth = d, M' = 4π(R-h)³d/3
And the mass of the earth M =4πR³d/3
M/M' = R³/(R-h)³
At the surface of the earth g =GM/R²
→g/g' = (GM/R²)/{GM'/(R-h)²}
→g/g' = M(R-h)²/M'R²
→g/g' = R/(R-h) =6400/(6400-1600) =6400/4800 =4/3
The time period of the pendulum in the mine T' =2π√(l/g')
The time period at the surface of the earth T = 2π√(l/g)
Hence T'/T =√(g/g') =√(4/3)
→T' = T*2/√3 ={2π√(0.40/9.8)}*2/√3
→T' = 1.27*2/1.732 s =1.47 s
31. A uniform plate of mass M stays horizontally and symmetrically on two wheels rotating in opposite directions (Figure 12-E16). The separation between the wheels is L. The friction coefficient between each wheel and the plate is µ. Find the time period of oscillation of the plate if it is slightly displaced along its length and released.
 |
| Figure for Q-31 |
ANSWER: When the center of the plate is above the midpoint of the line joining the wheels, both wheels get equal weight and the friction force by both the wheels are equal. Let the slight displacement along the length be d. If the normal force by the wheels are R and R', then
R+R' =Mg
and R*L = Mg*(d+L/2) {Taking torque about the right wheel}
→R =½Mg(2d+L)/L
 |
| Figure for Q-31 |
And R' = Mg-R
=Mg - ½Mg(2d+L)/L
=½[2MgL-MgL-2Mgd]/L
=½[MgL-2Mgd]/L
So net friction force on the plate F = µ(R-R')
=(µ/2L)[2Mgd+MgL-MgL+2Mgd]
=(µ/2L)[4Mgd]
=2µMgd/L
→F = M(2µg/L)d
Which is in the form of F =m⍵²r, ie. the force is proportional to the displacement. So it is a simple harmonic motion with
⍵ = √(2µg/L)
Hence the time period T = 2π/⍵ =2π√(L/2µg)
32. A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where g = π² m/s².
ANSWER: Time period T =2 s. But the time period of a simple pendulum is given by T = 2π√(l/g).
→l/g = T²/4π²
→l = gT²/4π²
putting the values,
l = π²*4/4π² = 1 m
33. The angle made by the string of a simple pendulum with the vertical depends on time as θ = (π/90).sin[(π s⁻¹)t]. Find the length of the pendulum if g = π² m/s².
ANSWER: The given equation θ = (π/90).sin[(π s⁻¹)t]
is in the form of the simple harmonic motion displacement A.sin⍵t. From comparison,
⍵ = π s⁻¹.
Time period T = 2π/⍵ =2π/π = 2 s
But for the simple pendulum T =2π√(l/g)
→2 = 2π√(l/π²) =2√l
→l = 1 m
34. The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours?
ANSWER: The pendulum of a clock has a correct time period of 2 s but this clock takes more time in one oscillation. So it is running slow. Each second it is slower by 2.04/2 -1 =0.02 s. Hence in 24 hr i.e. 24*3600 s it runs 24*3600*0.02 s =0.48*3600/3600 hr =0.48*60 minutes =28.80 minutes slow.
35. A pendulum clock giving correct time at a place where g = 9.800 m/s² is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.
ANSWER: The pendulum loses 24 s in 24 hrs, i.e. 1 s each hr or 3600 s. The correct clock has a time period T, of 2 s which makes 3600/2 =1800 oscillations per hour. But this clock is taking 3600+1 =3601 s in 1800 oscillations (In one hour). So the new time period T' = 3601/1800 s
If the value of g at the new place = g'
then, T' = 2π√(L/g') and T =2π√(L/g), dividing
T'/T = √(g/g')
→g' =gT²/T'² =9.80*2²/(3601/1800)²
=9.80*4/(2.0005)²
=9.795m/s²
36. A simple pendulum is constructed by hanging a heavy ball by 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system the system is taken on the moon where the acceleration due to the gravitation of the moon is 1.67 m/s²?
ANSWER: (a) l = 5.0 m, So T =2π√(l/g) =2π√(5/9.8) =1.429π seconds.
Oscillations per seconds = frequency =1/T = 1/1.429π Hz
=0.70/π Hz or oscillations per second.
(b) At moon g =1.67 m/s²
The frequency = 1/T= 1/{2π√(l/g)}
=1/{2π√(5/1.67)}
=1/(2π√3) Hz
37. The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.
ANSWER: The maximum tension will be when the string is vertical where it balances the weight as well as the centrifugal force. If the speed of the bob at this position is v, Tₘₐₓ = mg+mv²/l
The minimum tension in the string will be at the highest point of the oscillation where the speed of the bob is zero. Tₘᵢₙ =mg.cosθ.
|
| Figure for Q-37 |
Given that Tₘₐₓ = 2*Tₘᵢₙ
→mg+mv²/l = 2mg.cosθ
→cosθ = ½(1+v²/gl) .... (i)
Since in a SHM the total energy is constant at any instant, equating the total energies at these two positions we have,
½mv² = mg(l-l.cosθ) .........{Taking reference for P.E. at the lowest point of the bob}
→v² = 2gl(1-cosθ)
Putting in (i)
cosθ = ½{1+2(1-cosθ)}=½(3-2.cosθ) =3/2 - cosθ
→2.cosθ =3/2
→cosθ =3/4
→θ =cos⁻¹(3/4)
38. A small block oscillates back and forth on a smooth concave surface of radius R (figure 12-E17). Find the time period of small oscillation.
 |
| The figure for Q - 38 |
ANSWER: Let the mass of the mall block = m,
The normal force on the block = mg.cosθ where θ is the angle between the radius at any instant t with the vertical. The tangential force F = mg.sinθ
Acceleration of the block = a = mg.sinθ/m = g.sinθ
If the angular displacement θ is small then sinθ= θ and the block moves in approximately a straight line.
Now acceleration a =g.sinθ =gθ
If the displacement of the block from the lowest position is x, then θ=x/R
So, a =gx/R =(g/R)x
g/R is constant, hence the acceleration is proportional to the displacement. Thus it is a simple harmonic motion with ⍵² =g/R
→⍵ = √(g/R)
The time period of the oscillation =2π/⍵ =2π√(R/g)
39. A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillation about the lowest point. Find the time period.
ANSWER: Let the ball be displaced a distance x from the mean position. If the radius of the surface at this place makes an angle θ with the vertical, the restoring force on the ball = mg.sinθ
Since the ball rolls on the surface without slipping the torque of this force about the point of contact = mgr.sinθ
Moment of inertia of the ball about an axis passing through its center = 2mr²/5
M.I. of the ball about the axis passing through the point of contact
I = 2mr²/5 +mr² =7mr²/5
M.I. of the ball about the axis passing through the point of contact
I = 2mr²/5 +mr² =7mr²/5
Angular acceleration α = Torque/ M.I.
=mgr.sinθ/(7mr²/5)
=mgr.sinθ/(7mr²/5)
=5g.sinθ/7r
Linear acceleration of the center of ball, a=αr =5g.sinθ/7 ---------(ii)
If the angle θ is small, sinθ =θ =x/(R-r)
so, a = 5gx/7(R-r) ={5g/7(R-r)}x
Hence the acceleration is proportional to the displacement and it is in simple harmonic motion with ⍵² ={5g/7(R-r)}
Hence the time period T =2π/⍵ =2π√{7(R-r)/5g}40. A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep, calculate the time period of the pendulum there. Radius of the earth = 6400 km.
ANSWER: R = 6400 km
h = 1600 km, l=0.40 m
The gravitational field at the depth of 1600 km =GM'/(R-h)²
M' is the mass of the earth part with radius R-h.
The Force on the bob of the pendulum if its mass is m
= GM'm/(R-h)²
If the acceleration due to the gravity there =g'
Then mg' =GM'm/(R-h)²
g' = GM'/(R-h)²
If the density of the earth = d, M' = 4π(R-h)³d/3
And the mass of the earth M =4πR³d/3
M/M' = R³/(R-h)³
At the surface of the earth g =GM/R²
→g/g' = (GM/R²)/{GM'/(R-h)²}
→g/g' = M(R-h)²/M'R²
→g/g' = R/(R-h) =6400/(6400-1600) =6400/4800 =4/3
The time period of the pendulum in the mine T' =2π√(l/g')
The time period at the surface of the earth T = 2π√(l/g)
Hence T'/T =√(g/g') =√(4/3)
→T' = T*2/√3 ={2π√(0.40/9.8)}*2/√3
→T' = 1.27*2/1.732 s =1.47 s
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
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
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HC Verma's Concepts of Physics, Chapter-7, Circular Motion
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HC Verma's Concepts of Physics, Chapter-6, Friction
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For more practice on problems on friction solve these "New Questions on Friction" .
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For more practice on problems on friction solve these "New Questions on Friction" .
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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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HC Verma's Concepts of Physics, Chapter-4, The Forces
"Questions for short Answers"
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Click here for "The Forces" - Exercises
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