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SIMPLE HARMONIC MOTION:--
EXERCISES Q1 TO Q10
1. A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t=0 it is at position x=5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4 s.
ANSWER: The equation for the displacement x at time t in SHM is given as x=A.sin(⍵t+ẟ) where A= amplitude which is 10 cm here. Given that at t=0, x=5 cm, hence putting them in equation we get
5 cm = (10 cm).sin(⍵*0+ẟ)
→sinẟ=½
→ẟ=π/6
The time period T = 2π/⍵
→⍵ = 2π/T =2π/6 s [Given T=6 s]
Now the equation x=A.sin(⍵t+ẟ) becomes
x = (10 cm) sin[(2π/6 s)t+π/6]
The magnitude of the acceleration is given as
a = -⍵².A.sin(⍵t+ẟ)
→a =-(2π/6)²*10.sin[(2π/6)*4+π/6]
→a =-(10π²/9).sin[9π/6]
→a =-10.97*sin(3π/2) =-10.97*(-1) ≈11 cm/s²
2. The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 m/s and 10 m/s² at a certain instant. Find the amplitude and the time period of the motion.
ANSWER: The position in an SHM is given as
x=A.sin(⍵t+ẟ)
→A.sin(⍵t+ẟ)=2 ..................... (1)
velocity v=A⍵cos(⍵t+ẟ)
→A⍵.cos(⍵t+ẟ) = 100 ................ (2)
Magnitude of the Acceleration a=⍵²x
→⍵²*2/100 = 10
→⍵ =100*5 =10√5 s⁻¹
Time period of the motion T =2π/⍵ =2π/10√5 =0.28 s
From (1)
sin(⍵t+ẟ)=2/A
→sin²(⍵t+ẟ)= 4/A²
From (2)
cos(⍵t+ẟ) =100/A⍵ =100/10√5A =10/√5A
→ cos²(⍵t+ẟ) =100/5A²
→1-sin²(⍵t+ẟ) =20/A²
→1-4/A²=20/A²
→4/A²+20/A² = 1
→24 =A²
→A² = 24
→A=√24 =4.9 cm
Note: Be careful about the units of different entities given in the problem.
3. A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and the potential energies equal?
ANSWER: Let the required distance from the mean position be x. The kinetic energy at a distance x from the mean position is given as E = ½m⍵²(A²-x²) and the potential energy E' = ½m⍵²x². The required condition is E = E'.
½m⍵²(A²-x²) = ½m⍵²x²
→A²-x² = x²
→2x² = A²
→x² = A²/2
→x = A/√2
Given A = 10 cm.
∴ x = 10/√2 =10√2/2 =5√2 cm
4. The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s². Find the position(s) of the particle when the speed is 8 cm/s.
ANSWER: The magnitude of the acceleration a= ⍵²x, maximum a will be at x =A.
aₘₐₓ=⍵²A
→A = aₘₐₓ/⍵²
→A = 50/⍵²
Since the maximum velocity at the mean position is given as vₘₐₓ=A⍵
∴ 10 = A⍵
→10 = 50/⍵
→⍵ =5 s⁻¹
And A= 50/25 = 2 cm
The position of the particle is now x=A.sin⍵t {Taking x=0 at t=0}
→x = 2.sin5t .............................. (i)
and the speed v = A⍵.cos⍵t
→v = 2*5.cos5t, the given speed is 8 cm /s
→8 =10.cos5t
→cos5t = 8/10 =0.8
∴sin5t = √(1-cos²5t)
→sin5t = √(1-0.64) =√0.36 = ±0.6
∴ for v = 8 cm/s we have
x =2*(±0.6) cm [from (i)]
→x = ±1.2 cm from the mean position.
5. A particle having mass 10 g oscillates according to the equation x=(2.0 cm).sin[(100 /s)t+π/6]. Find the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration at t=0.
ANSWER: The given equation is in the form of the equation of an SHM x =A.sin(⍵t+ẟ)
(a) Comparing both the equation we get
Amplitude A = 2.0 cm
⍵ = 100 s⁻¹
ẟ =π/6
The time period T =2π/⍵ =2π/100 = 6.28/100 s =0.0628 s ≈0.063 s
Let the spring constant be k, then
⍵ =√(k/m) [given m = 10 g =0.01 kg]
→100 = √(k/0.01) =√k/0.1
→√k = 100*0.1 = 10
→k = 100 N/m
(b) The position is given as x=(2.0 cm).sin[(100 s⁻¹)t+π/6]
At t = 0
x = 2.0*sinπ/6 =2.0*½ =1.0 cm
The velocity is given as v = A⍵.cos(⍵t+ẟ)
→ v=(2.0 cm)(100 s⁻¹).cos[(100 s⁻¹)t+π/6]
At t = 0,
v = 200*cosπ/6 =200*√3/2 =100√3 cm/s =100*1.73 cm/s
→v = 173 cm/s =1.73 m/s
Acceleration a = ⍵²x, at t = 0, x = 1.0 cm
∴a = (100 s⁻¹)²*1.0 cm/s² =10000 cm/s² = 100 m/s²
6. The equation of motion of a particle started at t=0 is given by x=5.sin(20t+π/3) where x is in centimeter and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed
ANSWER: (a) The velocity of the given motion will be given as
v = 5*20*cos(20t+π/3)
For the particle to come to rest v = 0, hence
5*20*cos(20t+π/3) = 0
→cos(20t+π/3) = 0 = cosπ/2
→20t+π/3 = π/2
→20t = π/2 - π/3 =π/6
→t = π/120 s
(b) The magnitude of the acceleration is given as
a= 5*20²*sin(20t+π/3), for a =0
5*20²*sin(20t+π/3) = 0
→sin(20t+π/3) = 0 = sin0 or sinπ
→20t +π/3 = 0 or π
→20t = -π/3 or (π-π/3)
→t = -π/60 or 2π/60 s
→t= -π/60 or π/30 s
Since the time should not be negative,
t = π/30 s
(c) Since the velocity of the motion is given as
v = 5*20*cos(20t+π/3)
the maximum velocity would be when the value of the cosine is maximum i.e. 1.
→cos(20t+π/3) = 1 = cos0 or cosπ
→20t+π/3 = 0 or π
Solving as above in (b) t = π/30 s
7. Consider a particle moving in simple harmonic motion according to the equation
x = 2.0 cos(50πt+tan⁻¹0.75)
where x is in centimeter and t in second. The motion is started at t=0.
(a) When does the particle come to rest for the first time?
(b) When does the acceleration have its maximum magnitude for the first time?
(c) When does the particle come to rest for the second time?
ANSWER: (a) Since x = 2.0 cos(50πt+tan⁻¹0.75)
velocity v = dx/dt = -2.0*50π*sin(50πt+tan⁻¹0.75)
For the particle to come to rest, v = 0
→ -2.0*50π*sin(50πt+tan⁻¹0.75) = 0
→ sin(50πt+tan⁻¹0.75) = sin0 or sinπ
→ 50πt+tan⁻¹0.75 = 0 or π
→t = -tan⁻¹0.75/50π or (π-tan⁻¹0.75)/50π
Since the time will not be negative,
t = (π-tan⁻¹0.75)/50π =(3.14-0.64)/(50*3.14) =0.016 s
→t = 1.6x10⁻² s
(b) The magnitude of the acceleration of the given motion is
a = 2.0*(50π)²*cos(50πt+tan⁻¹0.75)
For the magnitude of the acceleration to be maximum, the value of cosine factor should be maximum i.e. 1.
→cos(50πt+tan⁻¹0.75) = 1 = cos0 or cosπ
→50πt+tan⁻¹0.75 = 0 or π
Solving as in (a) above, we get t = 1.6x10⁻² s
(c) The particle comes to rest at extreme positions in an SHM. Since it comes to rest first time at t = 1.6x10⁻² s the second time it will come to rest will be at another extreme which will occur at T/2 time after the first rest. So the required time = 1.6x10⁻²+T/2
= 1.6x10⁻²+2π/2⍵
= 1.6x10⁻²+2π/(2*50π) [from the equation ⍵ = 50π]
= 1.6x10⁻²+0.02
= 1.6x10⁻²+2.0x10⁻² s
= 3.6x10⁻² s
8. Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change the value from half the amplitude to the amplitude.
ANSWER: Let the displacement of the particle in the SHM be described as x = A.sin⍵t [assuming at t=0, x=0]
For x = A/2
A.sin⍵t =A/2
→sin⍵t=½ =sinπ/6
→⍵t = π/6
→t = π/6⍵ ----------- (i)
For x = A
A.sin⍵t =A
→sin⍵t= 1 =sinπ/2
→⍵t = π/2
→t = π/2⍵
So the time taken for the displacement to change the value from half the amplitude to the amplitude =π/2⍵ - π/6⍵ =π/3⍵ =π/(3*2π/T) =T/6
Or, since the minimum time to reach the amplitude from the mean position =T/4
∴for the displacement to change the value from half the amplitude to the amplitude =T/4 -π/6⍵
= T/4 - πT/(6*2π)
=T/4-T/12
=2T/12
=T/6
9. The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.10 N/m. What mass should be attached to the spring?
ANSWER: The time period of a pendulum clock is 2 s.
The time period of a spring-mass system T' = 2π√(m/k) should also be 2 s if replacing the pendulum. Hence,
2π√(m/k) = 2
→m/k = 1/π²
→m =k/π²
→m =0.10/π²
→m =0.0101 kg ≈10 g
10. A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum i.e., a pendulum having frequency same as that of the block.
ANSWER: The extension of the spring = 2A =mg/k
→m/k=2A/g (Where A is the amplitude of the oscillation of the spring-mass system)
∴⍵=√(k/m) =√(g/2A) ---------------- (i)
Diagram for the problem-10
Let the common frequency be ν. So the time period T and ⍵ will also be the same.
For the simple pendulum
⍵ = √(g/l), equating it with (i)
√(g/2A) = √(g/l)
→l = 2A
So the length of the pendulum is equal to the extension of the spring.
===<<<O>>>===
Links to the Chapters
1. A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t=0 it is at position x=5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4 s.
ANSWER: The equation for the displacement x at time t in SHM is given as x=A.sin(⍵t+ẟ) where A= amplitude which is 10 cm here. Given that at t=0, x=5 cm, hence putting them in equation we get
5 cm = (10 cm).sin(⍵*0+ẟ)
→sinẟ=½
→ẟ=π/6
The time period T = 2π/⍵
→⍵ = 2π/T =2π/6 s [Given T=6 s]
Now the equation x=A.sin(⍵t+ẟ) becomes
x = (10 cm) sin[(2π/6 s)t+π/6]
The magnitude of the acceleration is given as
a = -⍵².A.sin(⍵t+ẟ)
→a =-(2π/6)²*10.sin[(2π/6)*4+π/6]
→a =-(10π²/9).sin[9π/6]
→a =-10.97*sin(3π/2) =-10.97*(-1) ≈11 cm/s² 2. The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 m/s and 10 m/s² at a certain instant. Find the amplitude and the time period of the motion.
ANSWER: The position in an SHM is given as
x=A.sin(⍵t+ẟ)
→A.sin(⍵t+ẟ)=2 ..................... (1)
velocity v=A⍵cos(⍵t+ẟ)
→A⍵.cos(⍵t+ẟ) = 100 ................ (2)
Magnitude of the Acceleration a=⍵²x
→⍵²*2/100 = 10
→⍵ =100*5 =10√5 s⁻¹
Time period of the motion T =2π/⍵ =2π/10√5 =0.28 s From (1)
sin(⍵t+ẟ)=2/A
→sin²(⍵t+ẟ)= 4/A²
From (2)
cos(⍵t+ẟ) =100/A⍵ =100/10√5A =10/√5A
→ cos²(⍵t+ẟ) =100/5A²
→1-sin²(⍵t+ẟ) =20/A²
→1-4/A²=20/A²
→4/A²+20/A² = 1
→24 =A²
→A² = 24
→A=√24 =4.9 cm
Note: Be careful about the units of different entities given in the problem.
3. A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and the potential energies equal?
ANSWER: Let the required distance from the mean position be x. The kinetic energy at a distance x from the mean position is given as E = ½m⍵²(A²-x²) and the potential energy E' = ½m⍵²x². The required condition is E = E'.
½m⍵²(A²-x²) = ½m⍵²x²
→A²-x² = x²
→2x² = A²
→x² = A²/2
→x = A/√2
Given A = 10 cm.
∴ x = 10/√2 =10√2/2 =5√2 cm
4. The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s². Find the position(s) of the particle when the speed is 8 cm/s.
ANSWER: The magnitude of the acceleration a= ⍵²x, maximum a will be at x =A.
aₘₐₓ=⍵²A
→A = aₘₐₓ/⍵²
→A = 50/⍵²
Since the maximum velocity at the mean position is given as vₘₐₓ=A⍵
∴ 10 = A⍵
→10 = 50/⍵
→⍵ =5 s⁻¹
And A= 50/25 = 2 cm
The position of the particle is now x=A.sin⍵t {Taking x=0 at t=0}
→x = 2.sin5t .............................. (i)
and the speed v = A⍵.cos⍵t
→v = 2*5.cos5t, the given speed is 8 cm /s
→8 =10.cos5t
→cos5t = 8/10 =0.8
∴sin5t = √(1-cos²5t)
→sin5t = √(1-0.64) =√0.36 = ±0.6
∴ for v = 8 cm/s we have
x =2*(±0.6) cm [from (i)]
→x = ±1.2 cm from the mean position.
5. A particle having mass 10 g oscillates according to the equation x=(2.0 cm).sin[(100 /s)t+π/6]. Find the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration at t=0.
ANSWER: The given equation is in the form of the equation of an SHM x =A.sin(⍵t+ẟ)
(a) Comparing both the equation we get
Amplitude A = 2.0 cm
⍵ = 100 s⁻¹
ẟ =π/6
The time period T =2π/⍵ =2π/100 = 6.28/100 s =0.0628 s ≈0.063 s
Let the spring constant be k, then
⍵ =√(k/m) [given m = 10 g =0.01 kg]
→100 = √(k/0.01) =√k/0.1
→√k = 100*0.1 = 10
→k = 100 N/m
(b) The position is given as x=(2.0 cm).sin[(100 s⁻¹)t+π/6]
At t = 0
x = 2.0*sinπ/6 =2.0*½ =1.0 cm
The velocity is given as v = A⍵.cos(⍵t+ẟ)
→ v=(2.0 cm)(100 s⁻¹).cos[(100 s⁻¹)t+π/6]
At t = 0,
v = 200*cosπ/6 =200*√3/2 =100√3 cm/s =100*1.73 cm/s→v = 173 cm/s =1.73 m/s
Acceleration a = ⍵²x, at t = 0, x = 1.0 cm
∴a = (100 s⁻¹)²*1.0 cm/s² =10000 cm/s² = 100 m/s²
6. The equation of motion of a particle started at t=0 is given by x=5.sin(20t+π/3) where x is in centimeter and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed
ANSWER: (a) The velocity of the given motion will be given as
v = 5*20*cos(20t+π/3)
For the particle to come to rest v = 0, hence
5*20*cos(20t+π/3) = 0
→cos(20t+π/3) = 0 = cosπ/2
→20t+π/3 = π/2
→20t = π/2 - π/3 =π/6
→t = π/120 s
(b) The magnitude of the acceleration is given as
a= 5*20²*sin(20t+π/3), for a =0
5*20²*sin(20t+π/3) = 0
→sin(20t+π/3) = 0 = sin0 or sinπ
→20t +π/3 = 0 or π
→20t = -π/3 or (π-π/3)
→t = -π/60 or 2π/60 s
→t= -π/60 or π/30 s
Since the time should not be negative,
t = π/30 s
(c) Since the velocity of the motion is given as
v = 5*20*cos(20t+π/3)
the maximum velocity would be when the value of the cosine is maximum i.e. 1.
→cos(20t+π/3) = 1 = cos0 or cosπ
→20t+π/3 = 0 or π
Solving as above in (b) t = π/30 s
7. Consider a particle moving in simple harmonic motion according to the equation
x = 2.0 cos(50πt+tan⁻¹0.75)
where x is in centimeter and t in second. The motion is started at t=0.
(a) When does the particle come to rest for the first time?
(b) When does the acceleration have its maximum magnitude for the first time?
(c) When does the particle come to rest for the second time?
ANSWER: (a) Since x = 2.0 cos(50πt+tan⁻¹0.75)
velocity v = dx/dt = -2.0*50π*sin(50πt+tan⁻¹0.75)
For the particle to come to rest, v = 0
→ -2.0*50π*sin(50πt+tan⁻¹0.75) = 0
→ sin(50πt+tan⁻¹0.75) = sin0 or sinπ
→ 50πt+tan⁻¹0.75 = 0 or π
→t = -tan⁻¹0.75/50π or (π-tan⁻¹0.75)/50π
Since the time will not be negative,
t = (π-tan⁻¹0.75)/50π =(3.14-0.64)/(50*3.14) =0.016 s
→t = 1.6x10⁻² s
(b) The magnitude of the acceleration of the given motion is
a = 2.0*(50π)²*cos(50πt+tan⁻¹0.75)
For the magnitude of the acceleration to be maximum, the value of cosine factor should be maximum i.e. 1.
→cos(50πt+tan⁻¹0.75) = 1 = cos0 or cosπ
→50πt+tan⁻¹0.75 = 0 or π
Solving as in (a) above, we get t = 1.6x10⁻² s
(c) The particle comes to rest at extreme positions in an SHM. Since it comes to rest first time at t = 1.6x10⁻² s the second time it will come to rest will be at another extreme which will occur at T/2 time after the first rest. So the required time = 1.6x10⁻²+T/2
= 1.6x10⁻²+2π/2⍵
= 1.6x10⁻²+2π/(2*50π) [from the equation ⍵ = 50π]
= 1.6x10⁻²+0.02
= 1.6x10⁻²+2.0x10⁻² s
= 3.6x10⁻² s
8. Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change the value from half the amplitude to the amplitude.
ANSWER: Let the displacement of the particle in the SHM be described as x = A.sin⍵t [assuming at t=0, x=0]
For x = A/2
A.sin⍵t =A/2
→sin⍵t=½ =sinπ/6
→⍵t = π/6
→t = π/6⍵ ----------- (i)
For x = A
A.sin⍵t =A
→sin⍵t= 1 =sinπ/2
→⍵t = π/2
→t = π/2⍵
So the time taken for the displacement to change the value from half the amplitude to the amplitude =π/2⍵ - π/6⍵ =π/3⍵ =π/(3*2π/T) =T/6
Or, since the minimum time to reach the amplitude from the mean position =T/4
∴for the displacement to change the value from half the amplitude to the amplitude =T/4 -π/6⍵
= T/4 - πT/(6*2π)
=T/4-T/12
=2T/12
=T/6
9. The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.10 N/m. What mass should be attached to the spring?
ANSWER: The time period of a pendulum clock is 2 s.
The time period of a spring-mass system T' = 2π√(m/k) should also be 2 s if replacing the pendulum. Hence,
2π√(m/k) = 2
→m/k = 1/π²
→m =k/π²
→m =0.10/π²
→m =0.0101 kg ≈10 g
10. A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum i.e., a pendulum having frequency same as that of the block.
ANSWER: The extension of the spring = 2A =mg/k
→m/k=2A/g (Where A is the amplitude of the oscillation of the spring-mass system)
∴⍵=√(k/m) =√(g/2A) ---------------- (i)
 |
| Diagram for the problem-10 |
Let the common frequency be ν. So the time period T and ⍵ will also be the same.
For the simple pendulum
⍵ = √(g/l), equating it with (i)
√(g/2A) = √(g/l)
→l = 2A
So the length of the pendulum is equal to the extension of the spring.
===<<<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
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" .
Click here for → Friction - OBJECTIVE-II
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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
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)
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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
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