Unit 1.1 Energy in Simple Harmonic Motion
Task:
(a) Determine velocity of the object at any position x when an object is in simple harmonic motion. (b) Explain the relationship between amplitude and energy possessed by the object.
Assumptions:
There is no friction in the system. The motion is sinusoidal and of a single frequency.
Consider a spring mass system in simple harmonic motion. As the mass oscillates back and forth the energy is continually interchanged between potential (E_{p}) and kinetic (E_{k}) energies. According to the Law of conservation of enery the total energy is constant as there is no friction or other losses in this system.
The object stops for an instant at the extremes where velocity = 0 and has maximum velocity v_{max }as it passes o (in either direction).
When the object is at rest the E_{k} = 0 and E_{p} = max
When the object is at maximum velocity E_{p} = 0 and E_{k} = max
Recall
E_{p} = Work done
Work done = Average Force.displacement
E_{p} = (^{(F + 0)}/_{2}*)*x = ½Fx
Substuting F = kx from Hooke's Law
E_{p} = (kx)x
E_{p} = ½kx^{2}
and E_{k} = ½mv^{2}
Now By the law of conservation of energy E_{t} = E_{p} + E_{k}
At -A, E_{t} = E_{p} + E_{k}
E_{t} = ½kA^{2} + 0
At -A, E_{t} = ½kA^{2} --- Eq1
At *O*, E_{t} = 0 + ½mv^{2}_{max}
At *O*, E_{t} = ½mv^{2}_{max} ----Eq2
Combining Eq1 and Eq2 we get ½kA^{2} = ½mv^{2}_{max}
=> v^{2}_{max} = (^{k}/_{m})A^{2} ----Eq3
v_{max} = √(^{k}/_{m})A ---- Eq4
At A, E_{t} = ½kA^{2} + 0
At *x*, E_{t} = ½k*x*^{2} + ½mv^{2}
½kA^{2} = ½k*x*^{2} + ½mv^{2}
=>
----- Eq5
Substituting Eq3 into Eq5 we get
----- Eq6
Conclusion:
(a) From Eq6 we see that at any position x, the object moves back and forth, so its velocity can be either in the + or - direction, but its magnitude depends only on its position x.
V_{max} when x = A or –A and v = 0 when x = 0.
(b) In Eq1 we see that E_{t} = ½kA^{2} so clearly the total energy is proportional to the amplitude of the wave.
Exercise 1.1.11 |