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x=a sin(wt+p), where p=phi, the phase angle. for eg, what is w, the angular frequency?

11 years ago

Dear Nishigandha,

The equation of SHM is the mathematical equation used to calculate the displacement, x, from the mean position of the particle. Here 'a' represents the amplitude of the oscillation, which being the maximum displacement the particle undergoes from the mean position.

**'w' is the angular velocity. The physical significance of angular velocity can be understood by observing the case of circular motion of a particle. Angular velocity is the rate of change of angle of the particle from the centre of the circle.**

**p=phi is the initial phase angle. The particle may not start with zero phase, so we've to add its initial phase in the equation of SHM.Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best !!!Regards,Askiitians ExpertsHimanshu Geed**

11 years ago

Simple Harmonic motion

An object moving along the x-axis is said to exhibit simple harmonic motion if its position as a function of time varies as

x(t) = x0 + Acos(wt+f).

The object oscillates about the equilibrium position x0. If we choose the origin of our coordinate system such that x0 = 0, then the displacement x from the equilibrium position as a function of time is given by

x(t) = Acos(wt+f).

A is the amplitude of the oscillation, i.e. the maximum displacement of the object from equilibrium, either in the positive or negative x-direction. Simple harmonic motion is repetitive. The period T is the time it takes the object to complete one oscillation and return to the starting position. The angular frequency w is given by w = 2p/T. The angular frequency is measured in radians per second. The inverse of the period is the frequency f = 1/T. The frequency f = 1/T = w/2p of the motion gives the number of complete oscillations per unit time. It is measured in units of Hertz, (1Hz = 1/s).

The velocity of the object as a function of time is given by

v(t) = dx(t)/dt = -wAsin(wt+f),

and the acceleration is given by

a(t) = dv(t)/dt = -w2Acos(wt+f) = -w2x.

The quantity f is called the phase constant. It is determined by the initial conditions of the motion. If at t = 0 the object has its maximum displacement in the positive x-direction, then f = 0, if it has its maximum displacement in the negative x-direction, then f = p. If at t = 0 the particle is moving through its equilibrium position with maximum velocity in the negative x-direction then f = p/2. The quantity wt + f is called the phase.

In the figure below position and velocity are plotted as a function of time for oscillatory motion with a period of 5s. The amplitude and the maximum velocity have arbitrary units. Position and velocity are out of phase. The velocity is zero at maximum displacement, and the displacement is zero at maximum speed.

For simple harmonic motion, the acceleration a = -w2x is proportional to the displacement, but in the opposite direction. Simple harmonic motion is accelerated motion. If an object exhibits simple harmonic motion, a force must be acting on the object. The force is

F = ma = -mw2x .

It obeys Hooke's law, F = -kx, with k = mw2.

F = ma = md2x/dt2 with F = -kx, leads to the second-order differential equation

d2x/dt2 = -(k/m)x.

We now know how to solve this equation. The solution is

x(t) = Acos(wt+f), with w2 = k/m.

The solution to a second-order differential equation includes two constants of integration. Here these constants are A and f. They are determined by the initial conditions of the problem.

The same equations have the same solutions. Whenever you encounter a differential equation of the form d2x/dt2 = -b2x, you know that the solution is x(t) = Acos(wt+f), with w = b.

The force exerted by a spring obeys Hooke's law. Assume that an object is attached to a spring, which is stretched or compressed. Then the spring exerts a force on the object. This force is proportional to the displacement of the spring from its equilibrium position and is in a direction opposite to the displacement.

F = -kx

Assume the spring is stretched a distance A and then released. The object attached to the spring accelerates.

a = -(k/m)x

It gains speed as it moves towards the equilibrium position because its acceleration is in the direction of its velocity. When it is at the equilibrium position, the acceleration is zero, but the object has kinetic energy. It overshoots the equilibrium position and starts slowing down, because the acceleration is now in a direction opposite to the direction of its velocity. Neglecting friction, it comes to a stop when the spring is compressed by a distance A and then accelerates back towards the equilibrium position. It again overshoots and comes to a stop at the initial position when the spring is stretched a distance A. The motion repeats. The object oscillates back and forth. It executes simple harmonic motion. The angular frequency of the motion is

,

the period is

,

and the frequency is

.

Problems:

A particle oscillates with simple harmonic motion, so that its displacement varies according to the expression x = (5cm)cos(2t+p/6) where x is in centimeters and t is in seconds. At t = 0 find

(a) the displacement of the particle,

(b) its velocity, and

(c) its acceleration.

(d) Find the period and amplitude of the motion.

Solution:

(a) The displacement as a function of time is x(t) = Acos(wt+f). Here w = 2/s, f = p/6, and A = 5cm. The displacement at t = 0 is x(0) = (5cm)cos(p/6) = 4.33cm.

(b) The velocity at t = 0 is v(0) = -w(5cm)sin(p/6) = -5cm/s.

(c) The acceleration at t = 0 is a(0) = -w2(5cm)cos(p/6) = -17.3cm/s2.

(d) The period of the motion is T = ps, and the amplitude is 5cm.

A 20 g particle moves in simple harmonic motion with a frequency of 3 oscillations per second and an amplitude of 5cm.

(a) Through what total distance does the particle move during one cycle of its motion?

(b) What is its maximum speed? Where does that occur?

(c) Find the maximum acceleration of the particle. Where in the motion does the maximum acceleration occur?

Solution:

(a) The total distance d the particle moves during one cycle is from x = -A to x = +A and back to x = -A, so d = 4A = 20cm.

(b) The maximum speed of the particle is

vmax = wA = 2pfA = 2p15cm/s = 0.94m/s.

The particle has maximum speed when it passes through the equilibrium position.

(c) The maximum acceleration of the particle is

amax = w2A = (2pf)2A = 17.8m/s2.

The particle has maximum acceleration at the turning points, where it has maximum displacement.

A 1kg mass attached to a spring of force constant 25N/m oscillates on a horizontal frictionless track. At t = 0 the mass is released from rest at x = -3cm, that is the spring is compressed by 3cm. Neglect the mass of the spring. Find

(a) The period of its motion,

(b) the maximum value of its speed and acceleration, and

(c) the displacement, velocity and acceleration as a function of time.

Solution:

(a) The period is T = 2pSQRT(m/k) = 2pSQRT(1s2/25) = 1.26s.

(b) The angular acceleration is w = SQRT(k/m) = 5/s.

The maximum speed is vmax = wA = 15cm/s.

The maximum acceleration of the particle is amax = w2A = 0.75m/s2.

x(t) = Acos(wt+f) = (3cm)cos((5/s)t+p) = -(3cm)cos((5/s)t),

v(t) = -wAsin(wt+f) = (15cm/s)sin((5/s)t),

a(t) = -w2Acos(wt+f) = (0.75m/s2)cos((5/s)t).

Assume a mass suspended from a vertical spring of spring constant k. In equilibrium the spring is stretched a distance x0 = mg/k. If the mass is displaced from equilibrium position downward and the spring is stretched an additional distance x, then the total force on the mass is mg-k(x0+x) = -kx directed towards the equilibrium position. If the mass is displaced upward by a distance x, then the total force on the mass is mg-k(x0-x) = kx, directed towards the equilibrium position. The mass will execute simple harmonic motion. The angular frequency w = SQRT(k/m) is the same for the mass oscillating on the spring in a vertical or horizontal position. The equilibrium length of the spring about which it oscillates is different for the vertical position and the horizontal position.

Assume an object attached to a spring exhibits simple harmonic motion. Let one end of the spring be attached to a wall and let the object move horizontally on a frictionless table.

What is the total energy of the object?

The object's kinetic energy is

K = (1/2)mv2 = (1/2)mw2A2sin2(wt+f),

Its potential energy is elastic potential energy. The elastic potential energy stored in a spring displaced a distance x from its equilibrium position is U=(1/2)kx2. The object's potential energy therefore is

U = (1/2)kx2 = (1/2)mw2x2 = (1/2)mw2A2cos2(wt+f).

The total mechanical energy of the object is

E = K+U = (1/2)mw2A2(sin2(wt+f)+cos2(wt+f)) = (1/2)mw2A2.

The energy E in the system is proportional to the square of the amplitude.

E = (1/2)kA2.

It is a continuously changing mixture of kinetic energy and potential energy.

For any object executing simple harmonic motion with angular frequency w, the restoring force F = -mw2x obeys Hooke's law, and therefore is a conservative force. We can define a potential energy U = (1/2) mw2x 2, and the total energy of the object is given by E = (1/2)mw2A2.

Problems:

A particle that hangs from a spring oscillates with an angular frequency of 2 rad/s. The spring is suspended from the ceiling of an elevator car and hangs motionless (relative to the car) as the car descends at a constant speed of 1.5 m/s. The car then suddenly stops. Neglect the mass of the spring.

(a) With what amplitude does the particle oscillate?

(b) What is the equation of motion for the particle? (Choose the upward direction to be positive.)

Solution:

(a) When traveling in the elevator at constant speed, the total force on the mass is zero. The force exerted by the spring is equal in magnitude to the gravitational force on the mass, the spring has the equilibrium length of a vertical spring. When the elevator suddenly stops, the end of the spring attached to the ceiling stops. The mass, however has momentum, p = mv, and therefore starts stretching the spring. It moves through the equilibrium position of the vertical spring with its maximum velocity vmax = 1.5m/s.

Its velocity as a function of time is v(t) = -wAsin(wt+f).

Since vmax = wA and w = 2/s, the amplitude of the amplitude of the oscillations is A = 0.75m.

(b) The equation of motion for the particle is d2x/dt2 = -(k/m)x = -w2x. Its solution is

x(t) = Acos(wt+f) = 0.75mcos((2/s)t+f).

If we choose the t = 0 to be the time the elevator stops and let the upward direction be positive, then x(0) = 0, and v(0) = -1.5m/s. We therefore need f to be p/2.

A mass-spring system oscillates with an amplitude of 3.5cm. If the force constant of the spring of 250N/m and the mass is 0.5kg, determine

(a) the mechanical energy of the system,

(b) the maximum speed of the mass, and

(c) the maximum acceleration

Solution:

(a) We have m = 0.5kg, A = 0.035m, k = 250N/m, w2 = k/m = 500/s2, w = 22.36/s.

The mechanical energy of the system is E = (1/2)kA2 = 0.153J.

(b) The maximum speed of the mass is vmax = wA = 0.78m/s.

(c) The maximum acceleration is amax = w2A = 17.5m/s2.

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