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# Linear SHM

Module by: Sunil Kumar Singh. E-mail the author

The subject matter of this module is linear SHM – harmonic motion along a straight line about the point of oscillation. There are various physical quantities associated with simple harmonic motion. Here, we intend to have a closer look at quantities associated with SHM like velocity, acceleration, work done, kinetic energy, potential energy and mechanical energy etc. For the sake of completeness, we shall also have a recap of concepts already discussed in earlier modules.

The SHM force relation “F = -kx” is a generic form of equation for linear SHM – not specific to block-spring system. In the case of block-spring system, “k” is the spring constant. This point is clarified to emphasize that relations that we shall be developing in this module applies to all linear SHM and not to a specific case.

Since displacement of SHM can be represented either in cosine or sine forms, depending where we start observing motion at t = 0. For someone, it is easier to visualize beginning of SHM, when particle is released from positive extreme. On the other hand, expression in sine form is convenient as particle is at the center of oscillation at t = 0. For this reason, some prefer sine representation.

The very fact that there are two ways to represent displacement may pose certain ambiguity or uncertainty in mind. We shall , therefore, strive to maintain complete independence of forms with the understanding that when it is cosine function, then starting reference is positive extreme and if it is sine function, then starting reference is center of oscillation. In order to illustrate flexibility, we shall be using “sine” expression of displacement in this module instead of cosine function, which has so far been used.

## Displacement

The displacement of the particle is given by :

x = A sin ω t + φ x = A sin ω t + φ

where “A” is the amplitude,"ω" is angular frequency, “φ” is the phase constant and “ωt + φ” is the phase. Clearly, displacement is periodic with respect to time as it is represented by bounded trigonometric function. The displacement “x” varies between “-A” and “A”.

## Velocity

The velocity of the particle as obtained from the solution of SHM equation is given by :

v = ω A 2 x 2 v = ω A 2 x 2

This is the relation of velocity of the particle with respect to displacement along the path of oscillation, bounded between “-ωA” and “ωA”. We can obtain a relation of velocity with respect to time by substituting expression of displacement “x” in the above equation :

v = ω A 2 x 2 = ω { A 2 A 2 sin 2 ω t + φ } = ω A cos ω t + φ v = ω A 2 x 2 = ω { A 2 A 2 sin 2 ω t + φ } = ω A cos ω t + φ

We can ,alternatively, deduce this expression by differentiating displacement, “x”, with respect to time :

v = x t = t A sin ω t + φ = ω A cos ω t + φ v = x t = t A sin ω t + φ = ω A cos ω t + φ

The variation of velocity with respect to time is sinusoidal and hence periodic. Here, we draw both displacement and velocity plots with respect to time in order to compare how velocity varies as particle is at different positions.

The upper figure is displacement – time plot, whereas lower figure is velocity – time plot. We observe following important points about variation of velocity :

• If displacement is sine function, then velocity function is cosine function and vice-versa.
• The range of velocity lies between “-ωA” and “ωA”.
• The velocity attains maximum value two times in a cycle at the center – (i) moving from negative to positive extreme and then (ii) moving from positive to negative extreme.
• The velocity at extreme positions is zero.

## Acceleration

The acceleration in linear motion is given as :

a = - ω 2 x = - k m x a = - ω 2 x = - k m x

Substituting for displacement “x”, we get an expression in variable time, “t” :

a = - ω 2 x = - ω 2 A sin ω t + φ a = - ω 2 x = - ω 2 A sin ω t + φ

We can obtain this relation also by differentiating displacement function twice or by differentiating velocity function once with respect to time. Few important points about the nature of acceleration should be kept in mind :

1: Acceleration changes its direction about point of oscillation. It is always directed towards the center whatever be the position of the particle executing SHM.

2: Acceleration linearly varies with negative of displacement. We have seen that force-displacement plot is a straight line. Hence, acceleration – displacement plot is also a straight line. It is positive when “x” is negative and it is negative when “x” is positive.

3: Nature of force with respect to time, however, is not linear. If we combine the expression of acceleration and displacement, then we have :

a = - ω 2 x = - ω 2 A sin ω t + φ a = - ω 2 x = - ω 2 A sin ω t + φ

Here, we draw both displacement and acceleration plots with respect to time in order to compare how acceleration varies as particle is at different positions.

The upper figure is displacement – time plot, whereas lower figure is acceleration – time plot. We observe following important points about variation of acceleration :

• If displacement is sine function, then acceleration function is also sine function, but with a negative sign.
• The range of acceleration lies between “ - ω 2 A - ω 2 A ” and “ ω 2 A ω 2 A ”.
• The acceleration attains maximum value at the extremes.
• The acceleration at the center is zero.

4: Since force is equal to product of mass and acceleration, F = ma, it is imperative that nature of force is similar to that of acceleration. It is given by :

F = m a = - k x = - m ω 2 x = - m ω 2 A sin ω t + φ F = m a = - k x = - m ω 2 x = - m ω 2 A sin ω t + φ

## Frequency, angular frequency and time period

The angular frequency is given by :

ω = k m = | a x | = | acceleration displacement | ω = k m = | a x | = | acceleration displacement |

We have used the fact " F = m a = - k x F = m a = - k x " to write different relations as above.

Time period is obtained from the defining relation :

T = 2 π ω = 2 π m k = 2 π | x a | = | displacement acceleration | T = 2 π ω = 2 π m k = 2 π | x a | = | displacement acceleration |

Frequency is obtained from the defining relation :

ν = 1 T = ω 2 π = 1 2 π k m = 1 2 π | a x | = | acceleration displacement | ν = 1 T = ω 2 π = 1 2 π k m = 1 2 π | a x | = | acceleration displacement |

## Kinetic energy

The instantaneous kinetic energy of oscillating particle is obtained from the defining equation of kinetic energy as :

K = 1 2 m v 2 = 1 2 m ω 2 A 2 x 2 = 1 2 k A 2 x 2 K = 1 2 m v 2 = 1 2 m ω 2 A 2 x 2 = 1 2 k A 2 x 2

The maximum value of KE corresponds to position when speed has maximum value. At x = 0,

K max = 1 2 m ω 2 A 2 0 2 = 1 2 k A 2 K max = 1 2 m ω 2 A 2 0 2 = 1 2 k A 2

The minimum value of KE corresponds to position when speed has minimum value. At x = A,

K min = 1 2 m ω 2 A 2 A 2 = 0 K min = 1 2 m ω 2 A 2 A 2 = 0

By substituting for “x” in the equation of kinetic energy, we get expression of kinetic energy in terms of variable time, “t” as :

K = 1 2 m ω 2 A 2 cos 2 ω t + φ K = 1 2 m ω 2 A 2 cos 2 ω t + φ

The kinetic energy – time plot is shown in the figure. We observe following important points about variation of kinetic energy :

1: The KE function is square of cosine function. It means that KE is always positive.

2: The time period of KE is half that of displacement. We know the trigonometric identity :

cos 2 x = 1 2 1 + cos 2 x cos 2 x = 1 2 1 + cos 2 x

Applying this trigonometric identity to the square of cosine term in the expression of kinetic energy as :

K = 1 4 m ω 2 A 2 { 1 + cos 2 ω t + φ } = 1 4 m ω 2 A 2 { 1 + cos 2 ω t + 2 φ } K = 1 4 m ω 2 A 2 { 1 + cos 2 ω t + φ } = 1 4 m ω 2 A 2 { 1 + cos 2 ω t + 2 φ }

Applying rules for finding time period, we know that period of function “kf(x)” is same as that of “f(x)”. Hence, period of “K” is same as that of “1 + cos (2ωt + 2φ)”. Also, we know that period of function “f(x) + a” is same as that of “f(x)”. Hence, period of “K” is same as that of “cos (2ωt + 2φ)”. Now, period of “f(ax±b)” is equal to period of “f(x)” divided by “|a|”. Hence, period of “K” is :

Period = 2 π 2 ω = π ω = T 2 Period = 2 π 2 ω = π ω = T 2

As time period of variation of kinetic energy is half, the frequency of “K” is twice that of displacement. For this reason, kinetic energy – time plot is denser than that of displacement – time plot.

## Potential energy

We recall that potential energy is an attribute of conservative force system. The first question that we need to answer is whether restoring force in SHM is a conservative force? One of the assumptions, which we made in the beginning, is that there is no dissipation of energy in SHM. It follows, then, that restoring force in SHM is a conservative force.

Second important point that we need to address is to determine a reference zero potential energy. We observe that force on the particle in SHM is zero at the center and as such serves to become the zero reference potential energy. Now, potential energy at a position “x” is equal to negative of the work done in taking the particle from reference point to position “x”.

U = - W = - F d x = - - k x d x = k x d x U = - W = - F d x = - - k x d x = k x d x

Integrating in the interval, we have :

U = k 0 x x d x = k [ x 2 2 ] 0 x = 1 2 k x 2 U = k 0 x x d x = k [ x 2 2 ] 0 x = 1 2 k x 2

Thus, instantaneous potential energy of oscillating particle is given as :

U = 1 2 k x 2 = 1 2 m ω 2 x 2 U = 1 2 k x 2 = 1 2 m ω 2 x 2

The maximum value of PE corresponds to position when speed is zero. At x = A,

U max = 1 2 k A 2 = 1 2 m ω 2 A 2 U max = 1 2 k A 2 = 1 2 m ω 2 A 2

The minimum value of PE corresponds to position when speed has maximum value. At x = 0,

U min = 1 2 k X 0 2 = 0 U min = 1 2 k X 0 2 = 0

By substituting for “x” in the equation of potential energy, we get expression of kinetic energy in terms of variable time, “t” as :

U = 1 2 k A 2 sin 2 ω t + φ = 1 2 m ω 2 A 2 sin 2 ω t + φ U = 1 2 k A 2 sin 2 ω t + φ = 1 2 m ω 2 A 2 sin 2 ω t + φ

The potential energy – time plot is shown in the figure. We observe following important points about variation of kinetic energy :

1: The KE function is square of sine function. It means that PE is always positive.

2: The time period of KE is half that of displacement. We have already proved the same in the case of kinetic energy. We can extend the reason in the case of potential energy as well :

Period = 2 π 2 ω = π ω = T 2 Period = 2 π 2 ω = π ω = T 2

As time period of variation is half, the frequency of “U” is twice that of displacement. For this reason, potential energy – time plot is denser than that of displacement – time plot.

## Mechanical energy

The basic requirement of SHM is that mechanical energy of the system is conserved. At any point or at any time of instant, the sum of potential and kinetic energy of the system in SHM is constant. This is substantiated by evaluating sum of two energies :

E = K + U E = K + U

Using expressions involving displacement, we have :

E = 1 2 m ω 2 A 2 x 2 + 1 2 m ω 2 x 2 = 1 2 m ω 2 A 2 E = 1 2 m ω 2 A 2 x 2 + 1 2 m ω 2 x 2 = 1 2 m ω 2 A 2

The plots of kinetic, potential and mechanical energy with respect to displacement are drawn in the figure. Note that the sum of kinetic and potential energy is always a constant, which is equal to the mechanical energy of the particle in SHM.

We can also obtain expression of mechanical energy, using time dependent expressions of kinetic and potential energy as :

E = 1 2 m ω 2 A 2 cos 2 ω t + φ + 1 2 m ω 2 A 2 sin 2 ω t + φ E = 1 2 m ω 2 A 2 cos 2 ω t + φ + 1 2 m ω 2 A 2 sin 2 ω t + φ

E = 1 2 m ω 2 A 2 { cos 2 ω t + φ + sin 2 ω t + φ } = 1 2 m ω 2 A 2 E = 1 2 m ω 2 A 2 { cos 2 ω t + φ + sin 2 ω t + φ } = 1 2 m ω 2 A 2

The mechanical energy – time plot is shown in the figure. We observe following important points about variation of energy with respect to time :

• Mechanical energy – time plot is a straight line parallel to time axis. This signifies that mechanical energy of particle in SHM is conserved.
• There is transformation of energy between kinetic and potential energy during SHM.
• At any instant, the sum of kinetic and potential energy is equal to 1 2 m ω 2 A 2 1 2 m ω 2 A 2 or 1 2 k A 2 1 2 k A 2 , which is equal to maximum values of either kinetic or potential energy.

## Example

Problem 1: The potential energy of an oscillating particle of mass “m” along straight line is given as :

U x = a + b x c 2 U x = a + b x c 2

The mechanical energy of the oscillating particle is “E”.

1. Determine whether oscillation is SHM?
2. If oscillation is SHM, then find amplitude and maximum kinetic energy.

Solution : If the motion is SHM, then restoring force is a conservative force. The potential energy is, then, defined such that :

U = - F x U = - F x

F = U x = - 2 b x c F = U x = - 2 b x c

In order to find the center of oscillation, we put F = 0.

F = - 2 b x - c = 0 x c = 0 x = c F = - 2 b x - c = 0 x c = 0 x = c

This means that particle is oscillating about point x = c. The displacement of the particle in that case is “x-c” – not “x”. This, in turn, means that force is proportional to negative of displacement, “x-c”. Hence, particle is executing SHM.

Alternatively, put y = x-c :

F = - 2 b y F = - 2 b y

This means that particle is executing SHM about y = 0. This means x-c = 0, which in turn, means that particle is executing SHM about x = c.

The mechanical energy is related to amplitude by the relation :

E = 1 2 m ω 2 A 2 E = 1 2 m ω 2 A 2

A = 2 E m ω 2 A = 2 E m ω 2

Now, m ω 2 = k = 2 b m ω 2 = k = 2 b . Hence,

A = 2 E 2 b = E b A = 2 E 2 b = E b

The potential energy is minimum at the center of oscillation i.e. when x = c. Putting this value in the expression of potential energy, we have :

U min = a + b c - c 2 = a U min = a + b c - c 2 = a

It is important to note that minimum value of potential energy need not be zero. Now, kinetic energy is maximum, when potential energy is minimum. Hence,

K max = E U min = E a K max = E U min = E a

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