Oscillation is a periodic motion of a body about a mean position.
Key Concepts:
Periodic Motion: A motion that repeats itself after a fixed interval of time is called periodic motion.
Oscillatory Motion:A motion in which a body moves to and fro about a mean position is called oscillatory motion.
Simple Harmonic Motion (SHM): A type of oscillatory motion in which the restoring force is directly proportional to the displacement from the mean position and is always directed towards the mean position.
Displacement, Velocity, and Acceleration in SHM: These quantities vary sinusoidally with time.
Time Period and Frequency: The time period is the time taken for one complete oscillation, and frequency is the number of oscillations per unit time.
Energy in SHM: The total mechanical energy of a particle in SHM remains constant and is the sum of its kinetic and potential energies.
Damped Oscillations: Oscillations with decreasing amplitude due to dissipative forces.
Forced Oscillations and Resonance: When an external periodic force is applied to an oscillating system, it is called forced oscillation.
By understanding these concepts, we can analyze various oscillatory phenomena, such as the motion of a pendulum, a spring-mass system, and sound waves.
1. Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Ans : Periodic motion** is characterized by a motion that repeats itself after a fixed interval of time.
(a) A swimmer completing a round trip: This is not periodic motion. While the swimmer returns to the starting point, the time taken for each trip might not be the same, and the motion doesn’t repeat in a fixed interval.
(b) A freely suspended bar magnet: When displaced and released, the magnet oscillates about its N-S direction. This motion repeats itself in a fixed interval, making it **periodic motion**.
(c) A hydrogen molecule rotating: The rotation of a molecule is a continuous, repetitive motion. It rotates about its center of mass in a fixed time interval, making it **periodic motion**.
(d) An arrow released from a bow: This is a one-time event. The arrow moves in a straight line and doesn’t repeat its motion, so it’s not periodic.
Therefore, options (b) and (c) represent periodic motion.
2. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a U-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Ans : *Simple Harmonic Motion (SHM):**
*Analyzing the given examples:
(a) Rotation of Earth: This is periodic motion but not SHM. While it’s repetitive, the restoring force isn’t directly proportional to the displacement.
(b) Oscillating mercury column: This is nearly SHM. The restoring force, due to gravity, is approximately proportional to the displacement from the equilibrium position for small displacements.
(c)Ball bearing in a bowl: This is nearly SHM. The restoring force, due to gravity, is approximately proportional to the displacement from the lowest point for small displacements.
(d) Vibrations of a polyatomic molecule:: This is periodic motion but not SHM. The restoring forces involved in molecular vibrations are complex and not strictly proportional to displacement.
Therefore, options (b) and (c) represent nearly simple harmonic motion.**
3. Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ?
Ans :Analyzing the Plots
Periodic motion is characterized by a motion that repeats itself after a fixed interval of time. Let’s analyze each plot:
Plot (a): This plot shows a linear increase in position with time. The motion is not repetitive, so it’s not periodic.
Plot (b): This plot shows a repetitive pattern where the particle oscillates back and forth. The time taken for one complete oscillation (from peak to trough and back to peak) is the period. In this case, the period appears to be 2 seconds.
Plot (c): This plot shows a non-repetitive pattern. The motion is not periodic.
Plot (d): This plot also shows a repetitive pattern, indicating periodic motion. The period here seems to be 4 seconds.
Conclusion:
Plots (b) and (d) represent periodic motion.
Period of motion in plot (b): 2 seconds
Period of motion in plot (d): 4 seconds
4. Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion?
Give period for each case of periodic motion (ω is any positive constant):
(a) sin ωt – cos ωt
(b) sin3 ωt
(c) 3 cos (π/4 – 2ωt)
(d) cos ωt + cos 3ωt + cos 5ωt
(e) exp (–ω 2 t 2 )
(f) 1 + ωt + ω 2 t 2
Ans : Let’s analyze each function:
Simple Harmonic Motion (SHM):
A function represents SHM if it is sinusoidal, i.e., of the form sin(ωt + φ) or cos(ωt + φ).
Periodic but Not Simple Harmonic:
A function is periodic if it repeats itself after a fixed interval of time, but it’s not necessarily sinusoidal.
Non-Periodic Motion:
A function is non-periodic if it doesn’t repeat itself after a fixed interval of time.
Now, let’s analyze each function:
(a) sin ωt – cos ωt = √2 sin(ωt – π/4)
This is a sinusoidal function, so it represents simple harmonic motion.
Period = 2π/ω
(b) sin 3ωt
This is also a sinusoidal function, but with a frequency three times that of (a).
Period = 2π/(3ω)
(c) 3 cos (π/4 – 2ωt) = 3 cos(2ωt – π/4)
This is a sinusoidal function with a different phase shift.
Period = 2π/(2ω) = π/ω
(d) cos ωt + cos 3ωt + cos 5ωt
This function is a superposition of sinusoidal functions with different frequencies. It’s periodic but not simple harmonic. The period is the LCM of the individual periods, which in this case is 2π/ω.
(e) exp(-ω²t²)
This function decays exponentially with time and doesn’t repeat itself. It’s non-periodic.
(f) 1 + ωt + ω²t²
This function is a quadratic function of time and doesn’t repeat itself. It’s non-periodic.
5. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Ans :The figure shows a particle undergoing simple harmonic motion (SHM) between points A and B, which are 10 cm apart. Point O is the equilibrium position. The direction from A to B is considered positive.
(a) At the endpoint A, where the particle reaches its maximum displacement, its velocity is zero. The acceleration and force acting on it point towards the mean position O and are positive.
(b) At the endpoint B, the other extreme position, the particle’s velocity is again zero. However, the acceleration and force now point towards O and are negative.
(c) As the particle passes through the midpoint O while moving towards A, its velocity reaches a maximum negative value. At this instant, both the acceleration and force acting on the particle are zero.
(d) When the particle is 2 cm from B, at point C, and moving towards A, its velocity is negative. The acceleration and force acting on it, directed towards O, are also negative.
(e) At point D, 3 cm from A, the particle is moving towards B with a positive velocity. The acceleration and force acting on it, directed towards O, are also positive.
(f) When the particle is 4 cm away from A and moving back towards A, its velocity is positive as it’s directed along BA. The acceleration and force, pointing towards OB, are also positive.
6. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = –200x 2
(c) a = –10x
(d) a = 100x 3
Ans : For a motion to be simple harmonic, the acceleration (a) must be directly proportional to the displacement (x) and oppositely directed. Mathematically, this can be represented as:
a = -kx
where k is a positive constant.
Let’s analyze each option:
(a) a = 0.7x: Here, acceleration is directly proportional to displacement, but the negative sign is missing. So, this is not simple harmonic motion.
(b) a = -200x²: Here, acceleration is not directly proportional to displacement. It’s proportional to the square of the displacement. So, this is not simple harmonic motion.
(c) a = -10x: This equation is in the form a = -kx, indicating that the acceleration is directly proportional to the displacement and oppositely directed. Hence, this represents simple harmonic motion.
(d) a = 100x³: Here, acceleration is not directly proportional to displacement. It scales with the cube of the displacement. So, this is not simple harmonic motion.
Therefore, only option (c) represents simple harmonic motion.
7. The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Ans : **Given:**
* x(t) = A cos(ωt + φ)
* Initial position (x₀) = 1 cm at t = 0
* Initial velocity (v₀) = ω cm/s
* Angular frequency (ω) = π s⁻¹
**Part 1: Finding A and φ**
Initial Position:
x(0) = A cos(φ) = 1 cm
**Initial Velocity:**
v(t) = dx/dt = -Aω sin(ωt + φ)
v(0) = -Aω sin(φ) = ω cm/s
From the first equation, cos(φ) = 1/A.
From the second equation, sin(φ) = -1.
Since sin²φ + cos²φ = 1, we get:
(1/A)² + (-1)² = 1
Solving for A, we get:
A = √2 cm
Substituting A = √2 in cos(φ) = 1/A, we get:
cos(φ) = 1/√2
This implies φ = ±π/4. Since sin(φ) = -1, we choose φ = -π/4.
Therefore, the equation of motion is:
x(t) = √2 cos(πt – π/4)
**Part 2: Using Sine Function**
x(t) = B sin(ωt + α)
Initial position:
B sin(α) = 1
Initial velocity:
Bω cos(α) = ω
Upon dividing the second equation by the first, we obtain:
tan(α) = 1
This implies α = π/4.
Substituting α = π/4 in B sin(α) = 1, we get:
B = √2
Therefore, the equation of motion using the sine function is:
x(t) = √2 sin(πt + π/4)
8.A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?
Ans :Understanding the Problem
We have a spring balance with a known scale length and maximum weight capacity. A body suspended from it oscillates with a given period.
Determine weight of the body.
Solution:
Step 1: Calculate the Spring Constant (k)
When the body is at its maximum extension, the spring force equals the weight of the body. We can use Hooke’s Law to find the spring constant:
F = kx
Where:
F = force (weight of the body)
k = spring constant
x= maximum extension (scale length)
Given:
Maximum weight (F) = 50 kg * 9.81 m/s² = 490.5 N
Maximum extension (x) = 20 cm = 0.2 m
So,
k = F/x
= 490.5 N / 0.2 m = 2452.5 N/m
Step 2: Calculate the Mass of the Body (m)
Given,
(T): time period of a mass-spring system
T = 2π√(m/k)
Squaring both sides and rearranging, we get:
m = (T² * k) / (4π²)
Given:
Time period (T) = 0.6 s
Substituting the values:
m = (0.6² * 2452.5) / (4π²) ≈ 22.36 kg
Step 3: Find Weight of the Body:
Weight =
acceleration due to gravity* mass
Weight = 22.36 kg * 9.81 m/s² ≈ 219.16 N
Therefore, the weight of the body is approximately 219.16 N ;Weight = 22.36 kg
9.A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
Fig. 13.19
Determine (i) the frequency of oscillations,
(ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.
Ans : a) The frequency of oscillations,
T = 2π√(m/k)
Where:
T is the period
m is the mass attached to the spring (3 kg)
k is the spring constant (1200 N/m)
Substituting the values:
T = 2π√(3 kg / 1200 N/m) ≈ 0.31 s
Therefore, the period of oscillation is approximately 0.31 seconds.
(b) The maximum acceleration occurs at the extreme points of the motion, where the displacement is maximum. The maximum displacement (A) is 2.0 cm or 0.02 m.
Determine maximum acceleration (a_max) by using following formula:
a_max = ω²A
Where:
ω= angular frequency,
given by ω = √(k/m)
Substituting the values:
ω = √(1200 N/m / 3 kg) = 20 rad/s
Therefore, the maximum acceleration is:
a_max = (20 rad/s)² * 0.02 m = 8 m/s²
(c) The maximum speed (v_max) occurs at the equilibrium position, where the displacement is zero. It can be calculated using:
v_max = ωA
Substituting the values:
v_max = 20 rad/s * 0.02 m = 0.4 m/s
Therefore, the maximum speed of the mass is 0.4 m/s.
10.In Exercise 13.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Ans :(a) As the time is measured from the equilibrium position, the displacement x is given by the equation:
x = a sin ωt = 2 sin 20t
(b) When the body reaches its maximum extension to the right, it is at the extreme right position. The initial phase at this point is π/2.
x = a sin (ωt + π/2) = a cos ωt = 2 cos 20t
(c) When the body reaches its maximum compression to the left, it is at the extreme left position. The initial phase at this point is 3π/2.
x = a sin (ωt + 3π/2) = -a cos ωt = -2 cos 20t
Note: The three functions have the same amplitude and frequency. They only differ in their initial phase.
11.Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Fig. 13.20
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Ans :1) Consider any point A on the reference circle in Figure (a). Draw a perpendicular BN from A to the x-axis.
If ∠POA is θ, then ∠OAM is θ – π.
The displacement x is given by:
x = -3 sin(2πt/T) = -3 sin(2πt/4)
This equation represents Simple Harmonic Motion (SHM).
2) Consider any point B on the reference circle in Figure (b). Draw a perpendicular BN from B to the x-axis.
Then ∠BON is θ = ωt.
The displacement x is given by:
x = -2 cos(2πt/T) = -2 cos(2πt/4)
This equation also represents Simple Harmonic Motion (SHM).
12. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
(a) x = –2 sin (3t + π/3)
(b) x = cos (π/6 – t)
(c) x = 3 sin (2πt + π/4)
(d) x = 2 cos πt
Ans :
13. Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force F. Fig. 13.21 (a) What is the maximum extension of the spring in the two cases ? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ?
Ans :Analyzing the Two Spring-Mass Systems
(a) Maximum Extension
Case (a):
A single mass m is attached to one end of the spring.
The spring constant is k.
The applied force is F.
Using Hooke’s Law, we can find the maximum extension:
F = kx
Therefore, the maximum extension, x_max, is:
x_max = F/k
Case (b):
Two masses m are attached to both ends of the spring.
Each end of the spring experiences a force F.
Since both ends of the spring are stretched by the same force F, the total extension of the spring will be twice the extension in each half.
Therefore, the maximum extension in each half is F/(2k), and the total maximum extension is:
x_max = 2 * (F/(2k)) = F/k
Conclusion:
The maximum extension of the spring is the same in both cases, F/k.
(b) Period of Oscillation
Case (a):
Given,
period of oscillation for a mass-spring system
T = 2π√(m/k)
Case (b):
In this case, each half of the spring acts like a spring with a spring constant of 2k (since the effective length is halved). Thus, the effective spring constant for the entire system is 2k.
The period of oscillation for this system is:
T = 2π√(m/(2k)) = √(1/2) * 2π√(m/k)
Conclusion:
The period of oscillation in case (a) is 2π√(m/k).
The period of oscillation in case (b) is √(1/2) * 2π√(m/k), which is √(1/2) times the period in case (a).
14 .The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed ?
Ans : Finding the Maximum Speed of the Piston
Understanding the Problem:
We’re given the stroke (twice the amplitude) of a piston undergoing simple harmonic motion (SHM) and its angular frequency.
Deteremine its maximum speed.
Solution:
1. Find the Amplitude:
The stroke is twice the amplitude:
Stroke = 2A
1.0 m = 2A
A = 0.5 m
2. Use the Formula for Maximum Speed in SHM:
Given,
maximum speed (v_max) in SHM
v_max = ωA
Where:
ω is the angular frequency (200 rad/min)
A is the amplitude (0.5 m)
3. Convert Angular Frequency to rad/s:
First, we need to convert the angular frequency from rad/min to rad/s:
ω = 200 rad/min * (1 min / 60 s) = 10/3 rad/s
4. Calculate the Maximum Speed:
Now, we can calculate the maximum speed:
v_max = (10/3 rad/s) * 0.5 m ≈ 1.67 m/s
Therefore, the maximum speed of the piston is approximately 1.67 m/s.
15. The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s–2)
Ans :Understanding the Problem:**
We’re given the time period of a pendulum on Earth and the acceleration due to gravity on the Moon. We need to find the time period of the same pendulum on the Moon.
**Solution:**
The time period of a simple pendulum is given by:
T = 2π√(L/g)
Where:
* T = time period
* L = length of the pendulum
* g = acceleration due to gravity
For a given pendulum, the length L remains constant. So, we can write the ratio of time periods for Earth (E) and Moon (M):
T_M / T_E = √(g_E / g_M)
Substituting the values:
T_M / 3.5 s = √(9.8 m/s² / 1.7 m/s²)
Solving for T_M:
T_M ≈ 8.0 s
Therefore, the time period of the pendulum on the Moon’s surface is approximately 8.0 seconds.
16. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?
Ans : When the car moves on a circular track, the pendulum experiences an additional centripetal acceleration towards the center of the circular path, along with the acceleration due to gravity.
The effective acceleration acting on the pendulum bob is given by:
g’ = √(g² + (v²/R)²)
where:
g is the acceleration due to gravity
v is the speed of the car
R is the radius of the circular track
The time period of a simple pendulum is given by:
T = 2π√(L/g’)
Substituting g’ in the formula for the time period:
T = 2π√(L / √(g² + (v²/R)²))
Therefore, the time period of the pendulum in this case is:
T = 2π√(L / √(g² + (v⁴/R²)
17. A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl . The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Ans :Initially, the cylinder is in equilibrium with a height y submerged in the liquid. This means the weight of the cylinder equals the buoyant force acting on it:
Ahρg = Ayρ₁g
When the cylinder is slightly depressed by a distance Δy and released, an additional buoyant force acts as a restoring force. This force is given by:
F = A(y+Δy)ρ₁g – Ayρ₁g = Aρ₁gΔy
Given,
acceleration, a, of the cylinder
a = F/m = Aρ₁gΔy / Ahρ = ρ₁gΔy / hρ.
The direction of this acceleration is opposite to the displacement Δy. Since the acceleration is proportional to the negative displacement, the motion of the cylinder is Simple Harmonic Motion (SHM).
The time period T of this SHM is:
T = 2π√(displacement/acceleration)
= 2π√(Δy/a)
= 2π√(hρ/ρ₁g)
18. One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Ans : Analyzing the U-Tube Oscillation
Understanding the Problem:
We have a U-tube filled with mercury. By applying suction to one arm, a pressure difference is created, causing a difference in the mercury levels in the two arms. When the suction is removed, the mercury column oscillates. We need to show that this oscillation is simple harmonic.
Key Points:
Restoring Force: When the mercury column is displaced from its equilibrium position, the pressure difference between the two arms creates a restoring force.
Simple Harmonic Motion (SHM): Simple Harmonic Motion (SHM) occurs when a restoring force is directly proportional to and oppositely directed to the displacement.
Mathematical Analysis:
Let:
h be the difference in the heights of the mercury columns
ρ be the density of mercury
A be the cross-sectional area of the U-tube
g be the acceleration due to gravity
The differential pressure between the two arms is expressed as:
ΔP = ρgh
This pressure difference creates a restoring force F on the mercury column:
F = AΔP = Aρgh
Since the volume of the displaced mercury is Ah, the mass of the displaced mercury is m = ρAh.
Hence, The restoring force may be expressed as:
F = -mg(h/h) = -mgx/h
where x is the displacement of the mercury column from its equilibrium position.
This equation is in the form of Hooke’s law, F = -kx, where k = mg/h.
Conclusion:
Since the restoring force is directly proportional to the displacement and acts in the opposite direction, the mercury column executes simple harmonic motion. The period of oscillation can be calculated using the formula for a mass-spring system:
T = 2π√(m/k) = 2π√(h/g)
