Energy in SHM potential energy in shm simple harmonic motion kinetic energy equals potential energy

know find derive Energy in SHM potential energy in shm simple harmonic motion kinetic energy equals potential energy derivation formula in physics ?

What is Simple Harmonic Motion ?

Simple harmonic motion (SHM) is the simplest kind of oscillatory motion in which a body, displaced from its stable equilibrium position, oscillates to and fro about the position when released. If the displacement (x) from the equilibrium position is small, the restoring force (F) acting on the body is given by

F = – k x

where k is a positive constant, known as the force constant. In the SI system k is expressed in newton per metre (N m^–1). The acceleration (a) of the body is given by

Thus acceleration (a) = – constant x displacement x. The resulting motion is called simple harmonic motion. Thus, a simple harmonic motion is a motion in which the acceleration (i) is proportional to the displacement from the equilibrium position and (ii) is directed towards the equilibrium position.

 Characteristics of SHM
The displacement x in SHM at time t is given by
x = A sin (ω t + Φ)

where the three constants A, ω and Φ characterize the SHM, i.e., they distinguish one SHM from another. A SHM can also be described by a cosine function as

We will use the sine function. A cosine function is equally valid.

(1) Amplitude The amplitude of SHM is the maximum (positive or negative) value of the displacement from the equilibrium position. Quantity A (which is the coefficient of the sine or cosine function) is the amplitude of SHM.

(2) Time period The smallest time interval during which the motion repeats itself is called time period or simply period of SHM. The angular frequency w is related to time period T as

Frequency v of SHM is the number of complete oscillations completed in 1 second.

Angular frequency ω is the coefficient of time t in the sine or cosine function.

(3) Phase The quantity (ω t + Φ) is called the phase of SHM at time t; it describes the state of motion at that instant. The quantity Φ is the phase at time t = 0 and is called the phase constant or initial phase or epoch of the SHM. The phase constant is the time independent term in the cosine or sine function.

Velocity and Acceleration in SHM

The velocity V or the particle in SHM is given by

The acceleration a of the particle in SHM is given by

Notice that, when the displacement is maximum, i.e., when x = |A|, the velocity is zero but the acceleration is maximum = |ω² A|. But when the displacement is zero (x = 0), the velocity is maximum = |ω A| and the acceleration is zero.

Energy in SHM

At any instant of time t, the kinetic energy of the oscillator is given by

At any displacement x from the equilibrium position,

At that instant t, the potential energy of the oscillator is given by

Thus, although the kinetic energy and potential energy of a simple harmonic oscillator both change with time t and displacement x, the total energy is independent of both x and t and hence remains constant. This is expected because friction has been neglected.

Figure below shows the variation K.E., P.E. and E with t for the case Φ = 0.

The energy of an oscillator may decrease with time not only by friction (damping) but also due to radiation. The oscllating body imparts periodic motion to the particles of the medium in which it oscillates, thus producing waves. For example, a tuning fork or a string produces sound waves in the medium which results in a decrease in energy.

Expressions for Time Period of Mass-Spring System

(1) Horizontal Oscillations of a Mass-Spring System Consider a block of mass m placed on a horizontal frictionless surface and attached to a spring of negligible mass and spring constant k as shown in below.

The block is pulled to the right by a small distance x from the equilibrium position and released. The restoring force on the block is F = – kx. The acceleration of the block is

Since a ∝ (–x), the motion of the block is simple harmonic. Comparing Eq. (i) with a = – ω² x, we get

(2) Vertical Oscillations of a Mass-spring System Consider a massless spring suspended from a support. [Fig. below (a)]. A block of mass m is attached at the lower end, as a result, the string extends by an amount d given by [Fig. below (b)].

F =kd

mg = kd

This is the equilibrium state of the system.

hen the body is pulled through a distance y from this position and released [Fig. above (c)], the restoring force is F = – ky and the acceleration of the block is

Hence the motion is simple harmonic whose time period is

Equation (i) determines k. Time period given by Eq. (ii) is the same as for horizontal oscillation. It depends only on m and k and is independent of gravity.

(3) Parallel Combination of Springs Figure below shows the equilibrium state of a block connected to two springs which are joined in parallel. If the block is pulled down through a distance x, the extension produced in each spring will be x. The restoring forces in the springs are

F1 = – k1 x and F2 = – k2 x.

Total restoring force F = F₁ + F₂ = – (k₁ + k₂)x = – (k_p) x where k_p = k₁ + k₂ is the effective force constant of the parallel combination. The time period is given by

Figure below shows the equilibrium state of a block connected to two springs which are joined in series. The block is pulled down by a distance x. Let x₁ and x₂ be the extensions produced in the springs. The restoring force in each spring will be the same equal to

(5) A Block Connected between Two Springs

Figure 10.7 shows the equilibrium state of a block connected between two springs

If the block is displaced through a distance x, say to the right, the spring k1 is extended by x and spring k2 is compressed by x so that the restoring force exerted by each spring on the block is in the same direction (along the left). If F1 and F2 are the restoring forces,  F1 = – k1x and F2 = – k2 x,

the total restoring force is F = – (k1 + k2)x = – kx where k = (k1 + k2) is the effective force constant of the system. The time period is

Expressions for Time Period of Some other Systems

(1) A Ball Oscillating in a Concave Mirror

A small spherical steel ball is placed a little away from the centre of a concave mirror whose radius of curvature is R. When the ball is released, it begins to oscillate about the centre. Place a small steel ball at A, a little away from the centre O of a concave mirror of radius of curvature R (= OC = AC) as shown in Fig. below. Let ∠AOC = θ. If m is the mass of the ball, its weight mg acts vertically downwards at A. This force is resolved into two rectangular components: mg cos θ (which is balanced by the reaction of the mirror) and mg sin θ (which provides the restoring force F). Thus

F = – mg sin θ = – mg θ (since θ is small, R being very large).

where force constant K = mg/R. Thus the motion is harmonic and the angular frequency is given by

(2) Oscillation of a Liquid in a U-tube

The column of the liquid is displaced through y by gently blowing into the tube (Fig. above). The columns exhibit vertical oscillations. Let L, A and ρ be respectively, the length of the liquid column, area of cross-section of the tube and density of the liquid. We shall neglect viscous effects. Since the right-hand side column is higher by 2y, with respect to the column on the left-hand side, the mass of this column of liquid is m = 2Aρ y. The restoring force (which is a gravitational force) is given by

F = – mg = – 2Aρgy = – Ky

where the force constant K = 2Aρg. The angular frequency of the harmonic oscillation is

where M = ρ AL is the total mass of the liquid in oscillation. Thus

The time period of oscillation is

It is interesting to note that the period of oscillation does not depend on the density of the liquid or the area of cross-section of the tube.

(3) Oscillation of Floating Vertical Cylindrical Body

A cylindrical piece of wood of height h and density r floats in a liquid of density ρ_t . The cylinder is depressed slightly and released. Let A be the cross-sectional area of the cylinder and M its mass. Fig. below (a) shows the static equilibrium, the weight of the cylinder being balanced by the weight of the liquid it displaces.

If the cylinder is depressed through a distance x, as shown in Fig. 10.13 above, the buoyant force on it increases by ρl Agx, because ρl Ax is the mass of the liquid displaced by dipping, g being the acceleration due to gravity. If viscous effects are neglected, the restoring force on the cork is given by

where K = ρt Ag. Since F ∝ – x, the motion of the cork is simple harmonic. The time period of the motion is

(4) A Simple Pendulum A simple pendulum consists of a massless inextensible string fixed at one end O and having a small bob at the other end (Fig. below). When the bob is displaced from equilibrium position A to a position B and released, the component mg cos θ of its weight balances with tension T and it returns under a restoring force

note : T is independent of the mass of the bob.

(5) A Compound Pendulum

A compound pendulum is a rigid body capable of oscillating about an axis. The pendulum consists of a rod of mass M and length L which is pivoted at O and carries a bob of mass m at the other end as shown in Fig. below. The rod is displaced and released.

The pendulum tends to return under the influence of a restoring torque

If I is the moment of inertia of the system about the axis passing through O and perpendicular to the plane of the rod, the angular acceleration is

since a ∝ – θ, the motion is simple harmonic whose angular frequency is (compare with α = – w² θ) given by

(6) Horizontal Oscillations of a Cylinder-Spring System

A solid cylinder of mass M and radius R is connected to a spring of force constant k as shown in Fig. below.

Case (a): Cylinder slips without rolling.

The total energy of the system is translational K.E. + P.E. If x is the instantaneous displacement and v the velocity of the cylinder, the total energy is

Case (b): Cylinder rolls without slipping
In this case, the total energy of the system is

This the general formula. For a solid cylinder, I = MR²/2 and then we get

(7) Vertical Oscillations Mass-spring System Connected by a Pulley

Case (a) The block is pulled out through a small distance x and released. If M is the mass of the pulley and R its radius, the total energy of the system is [Fig. below]

Case (b) In the system shown in Fig. below (b), the pulley is a circular disc of mass M and radius R. The string does not slip over the pulley and the spring is elastic and of negligible mass.

The equilibrium position is the position when the net force acting on it is zero (so that it has no translatory motion) and the torque acting on it is also zero (so that it has no rotatory motion). There will be no torque if the tensions in the two strings are equal. Let this tension be T. The extension in the spring will be

x₀ = T/k

For translational equilibrium of the pulley,

2T = mg

2 x₀ k = mg

x₀ = mg/2k

If the pulley is pulled down by a small distance x, the total length of string and the spring is 2x. Since the length of the string is constant, the extension in the spring is 2x. Total potential energy of the system is

(8) A Block Attached to Three Springs A block of mass m is connected to three identical springs as shown in Fig. below. The block is pushed towards spring 1 through a small distance so that spring 1 is compressed by x and springs 2 and 3 are extended by x2 = x3 = x cos 45° = x / √2 . When the block is released, the restoring force acting on it is

Damped Oscillations

Oscillations under the influence of damping (or frictional) force are called damped oscillations. Due to damping the amplitude (and hence energy) of the oscillator keeps on decreasing with time and eventually the oscillator comes to rest. Damping also decreases the frequency of the oscillator.

Forced Oscillations and Resonance

The oscillations of a system under the influence of an external periodic force are called forced oscillations. The external force maintains the oscillations of a damped oscillator. The amplitude of these oscillations remains constant. If the frequency of the externally applied force is equal to the natural frequency of the oscillator, resonance is said to occur. If damping is small, the amplitude of resonant oscillations will become very large. At resonance, the oscillator absorbs maximum energy supplied by the external force.