English
Tamil Nadu Board of Secondary EducationHSC Science Class 11

Discuss in detail the energy in simple harmonic motion.

Advertisements
Advertisements

Question

Discuss in detail the energy in simple harmonic motion.

Long Answer
Advertisements

Solution

Energy in simple harmonic motion:

(a) Expression for Potential Energy: For the simple harmonic motion, the force and the displacement are related by Hooke’s law

`vec"F" = -"k"vec("r")`

Since force is a vector quantity, in three dimensions it has three components.

Further, the force in the above equation is a conservative force field; such a force can be derived from a scalar function which has only one component. In one dimensional case

F = − kx .....................(1)

The work done by the conservative force field is independent of the path. The potential energy U can be calculated from the following expression.

F = `-"dU"/"dx"` ..........(2)

Comparing (1) and (2), we get

`-"dU"/"dx" = -"kx"`

dU = kx dx

This work done by the force F during a small displacement dx stores as potential energy

U(x) = `int_0^"x" "kx′" "dx′" = 1/2["k"("x′")^2]_0^"x" = 1/2"kx"^2` ..........(3)

From equation `ω = sqrt("k"/"m")`, we can substitute the value of force constant k = mω2 in equation (3),

U(x) = `1/2"m"ω^2"x"^2` ...................(4)

where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation y = A sin ωt, we get

x = A sin ωt

U(t) = `1/2"m"ω^2"A"^2 sin^2ω"t"` ............(5)

This variation of U is shown in figure.


    Variation of potential energy with time t

(b) Expression for Kinetic Energy:

Kinetic energy KE = `1/2"mv"_"x"^2 = 1/2"m" ("dx"/"dt")^2` ............(6)

Since the particle is executing simple harmonic motion, from equation y = A sin ωt

x = A sin ωt

Therefore, velocity is

vx = `"dx"/"dt"` = Aω cos ωt ...................(7)

= `"A"ω sqrt(1 - ("x"/"A")^2)`

vx = `ω sqrt("A"^2 - "x"^2)`

Hence, KE = `1/2"mv"_"x"^2 = 1/2"m"ω^2 ("A"^2 - "x"^2)` ........(9)

KE = `1/2"m"ω^2 "A"^2 cos^2 ω"t"` ..............(10)

This variation with time is shown in figure.


    Variation of kinetic energy with time t

(c) Expression for Total Energy: Total energy is the sum of kinetic energy and potential energy

E = KE + U ..........…(11)

E = `1/2"m"ω^2 ("A"^2 - "x"^2) + 1/2 "m"ω^2 "x"^2`

Hence, cancelling x2 term,

E = `1/2"m"ω^2"A"^2` = constant ...............(12)

Alternatively, from equation (5) and equation (10), we get the total energy as

E = `1/2"m"ω^2"A"^2 sin^2 ω"t" + 1/2"m"ω^2"A"^2 cos^2 ω"t"`

= `1/2"m"ω^2"A"^2 (sin^2 ω"t" + cos^2 ω"t")`

From trigonometry identity, (sin2 ωt + cos2 ωt) = 1

E = `1/2"m"ω^2"A"^2` = constant

which gives the law of conservation of total energy. This is depicted in Figure


Both kinetic energy and potential energy vary but total energy is constant

Thus the amplitude of a simple harmonic oscillator can be expressed in terms of total energy.

A = `sqrt((2"E")/("m"ω^2)) = sqrt((2"E")/"k")` ..........(13)

shaalaa.com
Energy in Simple Harmonic Motion
  Is there an error in this question or solution?
Chapter 10: Oscillations - Evaluation [Page 220]

APPEARS IN

Samacheer Kalvi Physics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 10 Oscillations
Evaluation | Q III. 9. | Page 220

RELATED QUESTIONS

A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [(100 s−1)t + π/6]. Find (a) the amplitude, the time period and the spring constant. (c) the position, the velocity and the acceleration at t = 0.


Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 πt + tan−1 0.75) where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the first time? (b) When does he acceleration have its maximum magnitude for the first time? (c) When does the particle come to rest for the second time ?


The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.


Repeat the previous exercise if the angle between each pair of springs is 120° initially.


When a particle executing S.H.M oscillates with a frequency v, then the kinetic energy of the particle? 


If a body is executing simple harmonic motion and its current displacements is `sqrt3/2` times the amplitude from its mean position, then the ratio between potential energy and kinetic energy is:


Displacement versus time curve for a particle executing S.H.M. is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.


Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.


Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.


A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = A sin`(pit)/90`. The ratio of kinetic to the potential energy of this particle at t = 210s will be ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×