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SIMPLE HARMONIC MOTION:--
EXERCISES Q21 TO Q30
21. The springs shown in the figure (12-E7) are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.
Figure for Q-21
ANSWER: First let us find out the equivalent spring constant. If the equivalent of k₂ and k₃ be k', let us apply a force P along k₂ and k₃. So, P/k' =P/k₂+P/k₃
→1/k'=1/k₂+1/k₃
→k' = k₂k₃/(k₂+k₃)
Now the equivalent spring constant of k' and k₁ will be the equivalent spring constant of the system of springs. Let it be k. If the block gets a displacement x, then
½kx² =½k₁x²+½k'x²
→k = k₁ + k' = k₁ + k₂k₃/(k₂+k₃)
→k = (k₁k₂+k₂k₃+k₃k₁)/(k₂+k₃)
The amplitude of the motion will be the maximum stretching under the force F. Hence the amplitude A = F/k
→A = F/[(k₁k₂+k₂k₃+k₃k₁)/(k₂+k₃)]
→A = F(k₂+k₃)/(k₁k₂+k₂k₃+k₃k₁)
The frequency of the motion 𝜈 =1/T
→𝜈 = ⍵/2π =(1/2π)√(k/M)
→𝜈 = (1/2π)√[(k₁k₂+k₂k₃+k₁k₃)/M(k₂+k₃)]
22. Find the elastic potential energy stored in each spring shown in the figure (12-E8), when the block is in equilibrium. Also, find the time period of vertical oscillation of the block.
Figure for Q-22
ANSWER: The force Mg is acting on all the springs.
The elongation of upper spring = Mg/k₁. The elastic P.E. stored in upper spring =½(k₁)(Mg/k₁)² =½M²g²/k₁
The elongation of middle spring = Mg/k₂. The elastic P.E. stored in middle spring =½(k₂)(Mg/k₂)² =½M²g²/k₂
The elongation of lower spring = Mg/k₃. The elastic P.E. stored in lower spring =½(k₃)(Mg/k₃)² =½M²g²/k₃
The total elongation in the system will be equal to the sum of the elongations in all the three springs. If k is the equivalent spring constant of the system, then
Mg/k = Mg/k₁ + Mg/k₂ + Mg/k₃
→1/k = 1/k₁ + 1/k₂ + 1/k₃
The time period of the vertical oscillation T =2π√(M/k)
→T = 2π√[M(1/k)]
→T = 2π√[M(1/k₁+1/k₂+1/k₃)]
23. The string, the spring and the pulley shown in figure (12-E9) are light. Find the time period of the mass m.
The figure for Q-23
ANSWER: The force on the spring is the weight of the mass m in the upward direction. The spring is like an inverted spring with mass m but gravitation in opposite direction. Hence its time period will be the same, T = 2π√(m/k).
24. Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.
ANSWER: Let us consider the angular momentum of the system about the center of the pulley when the mass m is at distance x down the equilibrium position and its velocity is v. If the force on the spring at this instant is F then torque of this force about the center of the pully = Fr.
The angular momentum = mvr + I⍵' = mvr + Iv/r = (mr + I/r)v
{⍵' is the angular velocity of the pulley and v =⍵'r}
Figure for Q-24
But The torque is also equal to the rate of change of angular momentum. So
Fr = (mr+I/r)dv/dt
→F=(m+I/r²)a
But also F =kx, so
(m+I/r²)a = kx
→a = {k/(m+I/r²)}x = ⍵²x
Hence ⍵² = k/(m+I/r²)
→⍵ = √{k/(m+I/r²)}
So the time period T = 2π/⍵ = 2π√{(m+I/r²)/k}
25. Consider the situation shown in figure (12-E10). Show that if the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period.
Figure for Q-25
ANSWER: Let the slight displacement be 2x. When released, there is no external force on the system so the center of mass (which is the mid-point between the two masses) remains static. From this static mid-point displacement of one mass =x. Considering the half-length of the spring (from mid-point to one end) the spring constant = 2k. The force on the mass F = 2kx. If the acceleration of the block is a, then F = ma =2kx
→a = (2k/m)x
So the acceleration is proportional to the displacement. Thus it is a Simple Harmonic Motion for which ⍵² = 2k/m.
→⍵ = √(2k/m)
The time-period T = 2π/⍵ =2π√(m/2k)
26. A rectangular plate of sides a and b is suspended from a ceiling by two parallel strings of length L each (figure 12-E11). The separation between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period.
Figure for Q-26
ANSWER: Let the slight displacement of the plate produces angle θ between the strings and the vertical. Due to similarity tension in each string will be same, say F. F=mg/2. Where m is the mass of the plate. The component of this tension in horizontal direction =F.sinθ,
Total horizontal force =2F.sinθ =2*mg/2*sinθ =mg.sinθ.
The acceleration of the plate
a =mg.sinθ/m =g.sinθ
Figure for Q-26
When the displacement is small, θ is also small. For small θ,
sinθ =θ, and θ =x/L {wher x is the horizontal displacement of the plate corresponding to θ}
Now, a = gθ = g*x/L =(g/L)*x
Since the acceleration is proportional to the displacement, it is a Simple Harmonic Motion for which ⍵² =g/L →⍵=√(g/L)
Time period T =2π/⍵ =2π√(L/g).
27. A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.
Figure for Q-27
ANSWER: The amplitude A =0.10 m, m = 1 kg, k = 100 N/m.
When the block is at the mean position, the force on it =0.
⍵ =√(k/m) =√(100/1) =10
Velocity v = A⍵ =0.10*10 = 1 m/s
When the 3 kg block is placed over it, let the combined velocity be V. From the linear momentum conservation principle,
(1+3)*V = 1*v =1*1 =1
→V = 1/4 m/s
Now ⍵ =√(k/M) = √(100/4) =√25 =5
Since at mean position, V =A⍵
→A = V/⍵ =1/(4*5) =1/20 m = 0.05 m = 5 cm
Frequency =1/T = ⍵/2π = 5/2π Hz
28. The left block in figure (12-E13) moves at a speed v towards the right block placed in equilibrium. All collisions to take place are elastic and the surfaces are frictionless. Show that the motions of the two blocks are periodic. Find the time period of theses periodic motions. Neglect the width of the blocks.
Figure for Q-28
ANSWER: Since all collisions are elastic the motions will go on without losing the energy. Let us find the time period of each block.
When the left block hits the right block, the left block comes to stop and the right block gains the velocity v (Since the velocity of approach = velocity of separation).
Hence the time taken by the right block to return back to point of collision = ½T = ½*2π√(m/k) = π√(m/k)
Now it hits the left block and transfers its velocity to the left block and itself stops. Since the same K.E. is returned back to the left block its velocity = v. It travels a distance 2L when it again comes back to the point of collision. In this time taken = 2L/v. After this again the same motion is repeated, i.e. it is a periodic motion. The time period of this periodic motion = π√(m/k)+2L/v.
29. Find the time period of the motion of the particle shown in figure (12-E14). Neglect the small effect of the bend near the bottom.
Figure for Q-29
ANSWER: Assuming no energy loss, the ball will rise to the same vertical height on the other plane so that the total energy (in the form of P.E. at the top) is conserved. Taking the horizontal floor as zero potential energy level, the ball will have the maximum K.E. at the bend. Therefore the velocity at coming down to bend and the velocity at the rise at the bend will be equal.
The time period will be the time taken in reaching the original position.
The length of the left plane = (10 cm).cosec45° =(0.10 m)*√2
=0.10√2 m.
If the time to reach the bend is t, then from s = ut+½gt²
0.10√2 = 0 +½(g*cos45°)t²
{component of g along the plane =g*cos45°}
→t ² = 0.20√2/(g/√2) =0.40/g
→t = √(0.40/g) =√0.04 =0.2 s
The velocity at the bend v = 0 + g*cos45°*0.2
=10*0.2/√2 =2/√2 = √2 {taking g =10 m/s²}
The component of g along the plane = g.cos30°
=g√3/2 =10*1.73/2 =10*0.87 =8.7 m/s²
If the time to reach the highest point on right slope is t', then
0 = √2-8.7*t'
→t' =√2/8.7 =1.414/8.7 =0.16 s
So the total time taken in travel from top of the left plane to top of the right plane =t + t' =0.2 + 0.16 =0.36 s
Hence the time period of the motion = 2(t +t') =2*0.36 s =0.72 s
30. All the surfaces shown in figure (12-E15) are frictionless. The mass of the car is M, that of the block is m and the spring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through a length x₀ when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and of the car as seen from the road. (b) Find the time period(s) of the two simple harmonic motions.
Figure for Q-30
ANSWER: (a) The spring applies a force F = kx₀ on the block. The same force is applied to the car in the opposite direction at the time of release. Since there is no external force on the system the center of mass will remain static. The block and the car will move in the opposite directions and will have the same time period.
Let the amplitude of the block be x and the amplitude of the car be y. Since they move in opposite direction x₀ =x+y.
Also due to the center of mass being static, mx = My
→x =My/m
Therefore, x + y = x₀ becomes
My/m + y = x₀
→(M+m)y =mx₀
→y = mx₀/(M+m) ...... amplitude of the car.
And x = Mmx₀/m(M+m)
→x = Mx₀/(M+m) ........amplitude of the block.
(b) The acceleration of the block = F/m = kx₀/m
but a =⍵²x
→kx₀/m =⍵²x
→⍵² = kx₀/mx =kx₀(M+m)/mMx₀
→⍵² = k(M+m)/mM
→⍵ = √{k(M+m)/mM}
Time period T =2π/⍵
→T =2π√{mM/k(M+m)}
Both the block and the car have the same time periods.
21. The springs shown in the figure (12-E7) are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.
 |
| Figure for Q-21 |
ANSWER: First let us find out the equivalent spring constant. If the equivalent of k₂ and k₃ be k', let us apply a force P along k₂ and k₃. So, P/k' =P/k₂+P/k₃
→1/k'=1/k₂+1/k₃
→k' = k₂k₃/(k₂+k₃)
Now the equivalent spring constant of k' and k₁ will be the equivalent spring constant of the system of springs. Let it be k. If the block gets a displacement x, then
½kx² =½k₁x²+½k'x²
→k = k₁ + k' = k₁ + k₂k₃/(k₂+k₃)
→k = (k₁k₂+k₂k₃+k₃k₁)/(k₂+k₃)
The amplitude of the motion will be the maximum stretching under the force F. Hence the amplitude A = F/k
→A = F/[(k₁k₂+k₂k₃+k₃k₁)/(k₂+k₃)]
→A = F(k₂+k₃)/(k₁k₂+k₂k₃+k₃k₁)
The frequency of the motion 𝜈 =1/T
→𝜈 = ⍵/2π =(1/2π)√(k/M)
→𝜈 = (1/2π)√[(k₁k₂+k₂k₃+k₁k₃)/M(k₂+k₃)]
22. Find the elastic potential energy stored in each spring shown in the figure (12-E8), when the block is in equilibrium. Also, find the time period of vertical oscillation of the block.
 |
| Figure for Q-22 |
ANSWER: The force Mg is acting on all the springs.
The elongation of upper spring = Mg/k₁. The elastic P.E. stored in upper spring =½(k₁)(Mg/k₁)² =½M²g²/k₁
The elongation of middle spring = Mg/k₂. The elastic P.E. stored in middle spring =½(k₂)(Mg/k₂)² =½M²g²/k₂
The elongation of lower spring = Mg/k₃. The elastic P.E. stored in lower spring =½(k₃)(Mg/k₃)² =½M²g²/k₃
The total elongation in the system will be equal to the sum of the elongations in all the three springs. If k is the equivalent spring constant of the system, then
Mg/k = Mg/k₁ + Mg/k₂ + Mg/k₃
→1/k = 1/k₁ + 1/k₂ + 1/k₃
The time period of the vertical oscillation T =2π√(M/k)
→T = 2π√[M(1/k)]
→T = 2π√[M(1/k₁+1/k₂+1/k₃)]
23. The string, the spring and the pulley shown in figure (12-E9) are light. Find the time period of the mass m.
 |
| The figure for Q-23 |
ANSWER: The force on the spring is the weight of the mass m in the upward direction. The spring is like an inverted spring with mass m but gravitation in opposite direction. Hence its time period will be the same, T = 2π√(m/k).
24. Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.
ANSWER: Let us consider the angular momentum of the system about the center of the pulley when the mass m is at distance x down the equilibrium position and its velocity is v. If the force on the spring at this instant is F then torque of this force about the center of the pully = Fr.
The angular momentum = mvr + I⍵' = mvr + Iv/r = (mr + I/r)v
{⍵' is the angular velocity of the pulley and v =⍵'r}
 |
| Figure for Q-24 |
But The torque is also equal to the rate of change of angular momentum. So
Fr = (mr+I/r)dv/dt
→F=(m+I/r²)a
But also F =kx, so
(m+I/r²)a = kx
→a = {k/(m+I/r²)}x = ⍵²x
Hence ⍵² = k/(m+I/r²)
→⍵ = √{k/(m+I/r²)}
So the time period T = 2π/⍵ = 2π√{(m+I/r²)/k}
25. Consider the situation shown in figure (12-E10). Show that if the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period.
 |
| Figure for Q-25 |
ANSWER: Let the slight displacement be 2x. When released, there is no external force on the system so the center of mass (which is the mid-point between the two masses) remains static. From this static mid-point displacement of one mass =x. Considering the half-length of the spring (from mid-point to one end) the spring constant = 2k. The force on the mass F = 2kx. If the acceleration of the block is a, then F = ma =2kx
→a = (2k/m)x
So the acceleration is proportional to the displacement. Thus it is a Simple Harmonic Motion for which ⍵² = 2k/m.
→⍵ = √(2k/m)
The time-period T = 2π/⍵ =2π√(m/2k)
26. A rectangular plate of sides a and b is suspended from a ceiling by two parallel strings of length L each (figure 12-E11). The separation between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period.
 |
| Figure for Q-26 |
ANSWER: Let the slight displacement of the plate produces angle θ between the strings and the vertical. Due to similarity tension in each string will be same, say F. F=mg/2. Where m is the mass of the plate. The component of this tension in horizontal direction =F.sinθ,
Total horizontal force =2F.sinθ =2*mg/2*sinθ =mg.sinθ.
The acceleration of the plate
a =mg.sinθ/m =g.sinθ
 |
| Figure for Q-26 |
When the displacement is small, θ is also small. For small θ,
sinθ =θ, and θ =x/L {wher x is the horizontal displacement of the plate corresponding to θ}
Now, a = gθ = g*x/L =(g/L)*x
Since the acceleration is proportional to the displacement, it is a Simple Harmonic Motion for which ⍵² =g/L →⍵=√(g/L)
Time period T =2π/⍵ =2π√(L/g).
27. A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.
 |
| Figure for Q-27 |
ANSWER: The amplitude A =0.10 m, m = 1 kg, k = 100 N/m.
When the block is at the mean position, the force on it =0.
⍵ =√(k/m) =√(100/1) =10
Velocity v = A⍵ =0.10*10 = 1 m/s
When the 3 kg block is placed over it, let the combined velocity be V. From the linear momentum conservation principle,
(1+3)*V = 1*v =1*1 =1
→V = 1/4 m/s
Now ⍵ =√(k/M) = √(100/4) =√25 =5
Since at mean position, V =A⍵
→A = V/⍵ =1/(4*5) =1/20 m = 0.05 m = 5 cmFrequency =1/T = ⍵/2π = 5/2π Hz
28. The left block in figure (12-E13) moves at a speed v towards the right block placed in equilibrium. All collisions to take place are elastic and the surfaces are frictionless. Show that the motions of the two blocks are periodic. Find the time period of theses periodic motions. Neglect the width of the blocks.
 |
| Figure for Q-28 |
ANSWER: Since all collisions are elastic the motions will go on without losing the energy. Let us find the time period of each block.
When the left block hits the right block, the left block comes to stop and the right block gains the velocity v (Since the velocity of approach = velocity of separation).
Hence the time taken by the right block to return back to point of collision = ½T = ½*2π√(m/k) = π√(m/k)
Now it hits the left block and transfers its velocity to the left block and itself stops. Since the same K.E. is returned back to the left block its velocity = v. It travels a distance 2L when it again comes back to the point of collision. In this time taken = 2L/v. After this again the same motion is repeated, i.e. it is a periodic motion. The time period of this periodic motion = π√(m/k)+2L/v.
29. Find the time period of the motion of the particle shown in figure (12-E14). Neglect the small effect of the bend near the bottom.
 |
| Figure for Q-29 |
ANSWER: Assuming no energy loss, the ball will rise to the same vertical height on the other plane so that the total energy (in the form of P.E. at the top) is conserved. Taking the horizontal floor as zero potential energy level, the ball will have the maximum K.E. at the bend. Therefore the velocity at coming down to bend and the velocity at the rise at the bend will be equal.
The time period will be the time taken in reaching the original position.
The length of the left plane = (10 cm).cosec45° =(0.10 m)*√2
=0.10√2 m.
If the time to reach the bend is t, then from s = ut+½gt²
0.10√2 = 0 +½(g*cos45°)t²
{component of g along the plane =g*cos45°}
→t ² = 0.20√2/(g/√2) =0.40/g
→t = √(0.40/g) =√0.04 =0.2 s
The velocity at the bend v = 0 + g*cos45°*0.2
=10*0.2/√2 =2/√2 = √2 {taking g =10 m/s²}
The component of g along the plane = g.cos30°
The component of g along the plane = g.cos30°
=g√3/2 =10*1.73/2 =10*0.87 =8.7 m/s²
If the time to reach the highest point on right slope is t', then
0 = √2-8.7*t'
→t' =√2/8.7 =1.414/8.7 =0.16 s
So the total time taken in travel from top of the left plane to top of the right plane =t + t' =0.2 + 0.16 =0.36 s
Hence the time period of the motion = 2(t +t') =2*0.36 s =0.72 s
30. All the surfaces shown in figure (12-E15) are frictionless. The mass of the car is M, that of the block is m and the spring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through a length x₀ when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and of the car as seen from the road. (b) Find the time period(s) of the two simple harmonic motions.
 |
| Figure for Q-30 |
ANSWER: (a) The spring applies a force F = kx₀ on the block. The same force is applied to the car in the opposite direction at the time of release. Since there is no external force on the system the center of mass will remain static. The block and the car will move in the opposite directions and will have the same time period.
Let the amplitude of the block be x and the amplitude of the car be y. Since they move in opposite direction x₀ =x+y.
Also due to the center of mass being static, mx = My
→x =My/m
Therefore, x + y = x₀ becomes
My/m + y = x₀
→(M+m)y =mx₀
→y = mx₀/(M+m) ...... amplitude of the car.
And x = Mmx₀/m(M+m)
→x = Mx₀/(M+m) ........amplitude of the block.(b) The acceleration of the block = F/m = kx₀/m
but a =⍵²x
→kx₀/m =⍵²x
→⍵² = kx₀/mx =kx₀(M+m)/mMx₀
→⍵² = k(M+m)/mM
→⍵ = √{k(M+m)/mM}
Time period T =2π/⍵
→T =2π√{mM/k(M+m)}
Both the block and the car have the same time periods.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 12 - Simple Harmonic Motion
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
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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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
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HC Verma's Concepts of Physics, Chapter-4, The Forces
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In Q.27 conservation of energy isn't applicable?
ReplyDeleteDear student, you can also solve it using the principle of conservation of energy.
Delete😅
ReplyDelete