Question:

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.

Answer:

(a) Velocity = -, Acceleration = +, Force = - (b) Velocity = +, Acceleration = -, Force = + (c) Velocity = -, Acceleration = +, Force = - (d) Velocity = -, Acceleration = +, Force = - (e) Velocity = +, Acceleration = -, Force = + (f) Velocity = -, Acceleration = +, Force = -

Question:

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 it its maximum speed ?

Answer:

1. Convert angular frequency to frequency: 200 rad/min = 200/2π rad/s Frequency = 200/2π Hz

2. Calculate the period of the motion: T = 1/f = 1/(200/2π) = π/100 s

3. Calculate the maximum speed: Vmax = 2πfA = 2π(200/2π)(1.0) = 200 m/s

Question:

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)

Answer:

Step 1: Identify the given values. Given: g on the surface of the moon = 1.7 m s^−2 and g on the surface of the earth = 9.8 m s^−2, Time period on the surface of the earth = 3.5 s

Step 2: Use the formula T = 2π√(l/g) to calculate the time period on the surface of the moon.

Step 3: Substitute the given values in the formula. T = 2π√(l/g) T = 2π√(l/1.7)

Step 4: Calculate the time period on the surface of the moon. T = 2π√(l/1.7) T = 2π√(3.5/1.7) T = 4.72 s

Question:

You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is places on it. Also the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Answer:

a) To determine the spring constant k, use the equation: k = (4π2m)/x2

m = mass of the automobile = 3000 kg x = amplitude of oscillation = 15 cm = 0.15 m

k = (4π2m)/x2 k = (4π2*3000)/(0.152) k = 1,072,000 N/m

b) To determine the damping constant b, use the equation: b = (4πm)/(ln(A2/A1))

m = mass of the automobile = 3000 kg A1 = initial amplitude of oscillation = 15 cm = 0.15 m A2 = amplitude of oscillation after one complete oscillation = 50% of A1 = 0.075 m

b = (4πm)/(ln(A2/A1)) b = (4π*3000)/(ln(0.075/0.15)) b = 5,843 Ns/m

Question:

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillation is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J=−αθ, where J is the restoring couple and θ the angle of twist).

Answer:

1. Calculate the moment of inertia of the disc: I = mr2 = (10 kg)(0.15 m)2 = 0.225 kg m2

2. Calculate the angular frequency of oscillation: ω = 2π/T = 2π/1.5 s = 4.19 rad/s

3. Calculate the torsional spring constant: α = J/θ = Iω2/θ = (0.225 kg m2)(4.19 rad/s)2/θ = 17.2 Nm/rad

Question:

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 ?

Answer:

Step 1: Determine the acceleration of the car.

Acceleration of the car = v2/R

Step 2: Determine the effective acceleration of the pendulum.

Effective acceleration of the pendulum = (v2/R) + g

Step 3: Calculate the time period of the pendulum.

Time period of the pendulum = 2π√(l/g + (v2/Rg))

Question:

Among the following which are the examples of periodic motion? (a) A hydrogen molecule rotating about its centre of mass. (b) A freely suspended bar magnet displaced form its N-S direction and released. (c) A swimmer completing one (return) trip from one bank of a river to the other and back. (d) An arrow release from a bow.

Answer:

Answer: (a) A hydrogen molecule rotating about its centre of mass. (b) A freely suspended bar magnet displaced form its N-S direction and released.

Question:

The motion of a particle executing simple harmonic motion is described by the displacement function, x(t)=Acos(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 πs^−1. If instead of the cosine function we choose the sine function to describe the SHM:x=Bsin(ωt+α), what are the amplitude and initial phase of the particle with the above initial conditions.

Answer:

Answer:

A) For the displacement function x(t)=Acos(ωt+ϕ):

Amplitude A = 1 cm Initial phase angle ϕ = 0

B) For the displacement function x(t)=Bsin(ωt+α):

Amplitude B = 1 cm Initial phase angle α = π/2

Question:

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) sinωt−cosωt (b) sin^3ωt (c) 3cos(π/4−2ωt) (d) cosωt+cos3ωt+cos5ωt (e) exp(−ω^2t^2) (f) 1+ωt+ω^2t^2

Answer:

(a) A: Simple Harmonic, Period: 2π/ω (b) B: Periodic but not Simple Harmonic, Period: 2π/ω (c) B: Periodic but not Simple Harmonic, Period: 4π/ω (d) B: Periodic but not Simple Harmonic, Period: 2π/ω (e) C: Non-periodic Motion (f) C: Non-periodic Motion

Question:

A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20cm. A body suspended from this balance, when displaced and released, oscillates with period of 0.6s. What is the weight of the body ?

Answer:

1. Calculate the acceleration due to gravity (g) using the formula g = 4π2/T2, where T is the period of oscillation.

g = 4π2/0.62 = 39.48 m/s2

2. Calculate the weight (W) of the body using the formula W = mg, where m is the mass of the body and g is the acceleration due to gravity.

W = mg = (50kg/20cm) x 39.48 m/s2 = 987.6 kg

Question:

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=−2sin(3t+π/3) (b) x=cos(π/6−t) (c) x=3sin(2πt+π/4) (d) x=2cosπt

Answer:

(a) Initial position: (-2, 0) Radius: 2 Angular speed: 3 rad/s

(b) Initial position: (1/2, sqrt(3)/2) Radius: 1 Angular speed: π/6 rad/s

(c) Initial position: (0, 3) Radius: 3 Angular speed: 2π rad/s

(d) Initial position: (2, 0) Radius: 2 Angular speed: π rad/s

Question:

Answer the following questions : (a) Time period of a particle in SHM depends on the force constant k and mass m of the particle: T=2π√m​/k ​. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum? (b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angle of oscillation a more involved analysis shows that T is greater than 2π√l​/g ​. Think of a qualitative argument to appreciate this result. (c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ? (d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ?

Answer:

(a) The time period of a pendulum is independent of the mass of the pendulum because the force constant k is dependent on the length of the pendulum, not the mass. Therefore, the time period is determined by the length of the pendulum, not the mass.

(b) As the angle of oscillation increases, the restoring force acting on the pendulum decreases, resulting in a longer time period.

(c) No, the watch will not give the correct time during free fall because the motion of the watch is not uniform due to the acceleration of gravity.

(d) The frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity is equal to the frequency of oscillation of a simple pendulum at rest on the ground. This is because the acceleration of gravity is the same for both situations.

Question:

The end which contain mercury of a U-tube 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.

Answer:

1. When the suction pump is removed, atmospheric pressure is equalized in both columns of the U-tube.

2. This causes the mercury in the U-tube to move in a direction that equalizes the pressure, resulting in a pressure difference between the two columns.

3. This pressure difference creates a restoring force that causes the mercury to oscillate, resulting in simple harmonic motion.

Question:

An air chamber of volume V has a neck of cross-sectional area a into which a light ball of mass m just fits and can move up and down without friction. The diameter of the ball is equal to that of the neck of the chamber. The ball is pressed down a little and released. If the bulk modulus of air is B, the time period of the oscillation of the ball is A T=2π√Ba^2​/mV ​B T=2π√BV​?ma^2 ​C T=2π√mB​/Va^2 ​D T=2π√mV/Ba^2

Answer:

Answer: D T=2π√mV/Ba^2

Question:

Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answer:

1. Recall that the kinetic energy of a particle in SHM is given by K=½mv² and the potential energy is given by U=½kx².

2. Consider the average kinetic energy over one period of oscillation. This is given by the integral of the kinetic energy over the period, divided by the period.

3. Similarly, consider the average potential energy over one period of oscillation. This is given by the integral of the potential energy over the period, divided by the period.

4. Show that the integrals of the kinetic and potential energies over the period are equal.

5. Since the integrals of the kinetic and potential energies are equal, it follows that the average kinetic energy over one period of oscillation equals the average potential energy over the same period.

Question:

A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0​ and pushed towards the centre with a velocity v0​ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω,x0​ and v0​. [Hint : Start with the equation x=acos(ωt+θ) and note that the initial velocity in negative.]

Answer:

1. Begin by writing the equation of motion for the mass attached to a spring: x = Acos(ωt + θ), where A is the amplitude, ω is the angular velocity, t is time and θ is the phase angle.

2. At time t = 0, the mass is pulled to a distance x0 and pushed towards the centre with a velocity v0. Therefore, x0 = Acos(θ) and v0 = -Aωsin(θ).

3. Solve for A by substituting x0 and v0 into the equation A = x0/cos(θ) and Aω = v0/sin(θ) and combining the two equations.

4. The amplitude of the resulting oscillations can then be determined in terms of the parameters ω, x0 and v0 as A = √(x0^2 + (v0/ω)^2).

Question:

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

Answer:

(a) Acceleration = -50 cm/s2, Velocity = 0 cm/s

(b) Acceleration = -30 cm/s2, Velocity = 10 cm/s

(c) Acceleration = 0 cm/s2, Velocity = 0 cm/s

Question:

A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1​. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T=2π√hρ​/ρ1​g where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer:

1. Begin by calculating the buoyancy force on the cork. This is equal to the weight of the liquid displaced by the cork, which is equal to the volume of the cork times the density of the liquid. The volume of the cork is equal to the base area times the height, so the buoyancy force is equal to Aρ1h.

2. Next, calculate the restoring force on the cork. This is equal to the weight of the cork minus the buoyancy force, which is equal to (Aρh - Aρ1h).

3. The cork will oscillate up and down with a period given by the equation T=2π√hρ/ρ1g. This equation comes from the fact that the restoring force is proportional to the displacement of the cork, and that the period of a simple harmonic oscillator is given by 2π√m/k, where m is the mass of the cork and k is the restoring force. Substituting the mass of the cork (ρh) and the restoring force (Aρh - Aρ1h) into the equation gives the desired result.

Question:

Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion ? (a) a = 0.7 x (b) a = -200x^2 (c) a = - 10x (d) a = 100x^3

Answer:

Answer: (c) a = - 10x