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Wize University Physics Textbook (Master) > Periodic Motion: Oscillations

Energy Conservation

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Energy of An Oscillator


There are two different type of energies related to an oscillation: Kinetic energy and potential energy
  • The kinetic energy is the energy of a moving object:
K =12mv2\boxed{K\ =\frac{1}{2}mv^2}
  • The potential energy is the energy stored in the oscillation. For mass and spring oscillation, it is a potential energy stored inside of the spring due to stretch or compression in the length of the spring:

U =12kx2\boxed{U\ =\frac{1}{2}kx^2}

Watch Out!
Kinetic energy is also shown by KEKE, EkE_kor UkU_k. Similarly, potential energy is also shown by PEPE, EpE_por UpU_p.

  • The total energy of the system at any time during SHM is the sum of kinetic energy and potential energy and is constant because of the conservation of energy:


E=K+U=constant\boxed{E=K+U= \text{constant}}



Watch Out!
Note that we have a seesaw behaviour between kinetic energy and potential energy. They can each change but sum of them should remain conserved!

Exam Tip
Since energy is conserved in a simple harmonic motion, if we know the total energy at a moment, we know it at any other moment as well!

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How to Find Total Energy of an Oscillator


We can find total energy of an oscillator using the three following methods:

  1. We can find it directly from its definition if we know position and velocity of the oscillator at any moment.

E=12m(v(t))2+12k(x(t))2\boxed{E=\frac{1}{2}m(v(t))^{2}+\frac{1}{2}k(x(t))^2}

  1. We can find it by looking at the equilibrium point.
At equilibrium: x = 0, and V = Vmax


E=K+U=12mvmax⁡2+0E=K+U=\frac{1}{2}mv_{\max^2}+0

E=12mA2ω2\boxed{E=\frac{1}{2}mA^2\omega^2}
  1. We can find it by looking at turning points.
At turning points: x=±Ax=\pm A and V=0

E =K+U=0+ 12kA2E\ =K+U=0+\ \frac{1}{2}kA^2

E=12kA2\boxed{E=\frac{1}{2}kA^2}


Exam Tip
Above equations are helpful in solving problems using conservation of energy:

12kA2=12kx2+12mv2\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2

Note: It is usually easier to solve problems using conservation of energy if the oscillator starts with a non-zero initial velocity.



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Example: Energy of an Oscillating Block


A block of unknown mass mmis attached to a horizontal spring which has a spring constant k=4.8 Nmk=4.8\ \frac{N}{m}. It is then set oscillating on a frictionless surface with an amplitude ofA=12 cmA=12\ cm. It takes 0.5s for the block to complete one full cycle of oscillation. Calculate:

a) The mass mm
b) The maximum acceleration amax⁡a_{\max}
c) The total energy of the system
d) At which point(s) the kinetic energy of the mass is 13\frac{1}{3}of the total energy of the system.

Solution:

Part a)

ω=2πT=2π0.5=12.567  rads\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.5}=12.567 \ \ \frac{rad}{s}

ω=km⇒m=kω2=0.0304 kg\omega=\sqrt{\frac{k}{m}}\Rightarrow m=\dfrac{k}{\omega^2}=0.0304\ kg

Part b)

amax⁡=Aω2=0.12(12.567)2=18.95 m/s2a_{\max}=A\omega^2=0.12\left(12.567\right)^2=18.95\ m/s^2

Part c)

E=12kA2=12(4.8)(0.12)2=0.0346 JE=\frac{1}{2}kA^2=\frac{1}{2}(4.8)(0.12)^2=0.0346 \ J

Part d)

{K=13E⇒23E=U=12kx2K+U=E\left\{\begin{array}{l}K=\frac{1}{3}E\Rightarrow\frac{2}{3}E=U=\frac{1}{2}kx^2\\K+U=E\end{array}\right.


⇒x2=43Ek=0.0096⇒x=±0.098 m \Rightarrow x^2=\dfrac{4}{3}\dfrac{E}{k}=0.0096\Rightarrow x=\pm0.098\ m\

A simple mass mm on a spring with spring constant kk oscillates about its equilibrium with amplitude AA. If the energy of the system suddenly decreases to half of the initial value, what will be the mass' new maximum speed?
Extra Practice
50 kgkg person taking a bungee jump:
a. If the initial length of the bungee cord is 20 mm and at the lowest point, the length of the cord is 25mm, what is the spring constant of the cord?
b. After the jump, the person is experiencing a damping oscillation. What is the damping constant if the amplitude of its oscillation reduces to 15\frac{1}{5} of its original amplitude after 20 s?
This question concerns the great garbage patch in the middle of Pacific ocean. Consider the garbage patch to be a perfect cylinder. At this moment 100 seagulls are sitting on the patch. Every time a wave hits the garbage patch the garbage patch bobs up and down 5 times in 7.2 seconds. When the garbage patch is still the 100 seagulls leave. You know notice that the garbage bobs up and down 5 times in 7 seconds.
The garbage patch weighs 1000 kg. (To make calculations simpler this number is much smaller than the real number.)
a) What is the total mass of the 100 seagulls
Conservation of energy: Spring
A spring when compressed 40 cm, has 20J of elastic potential energy. We cut the spring in half. We compress one half for 30 cm. How much elastic energy is stored in the half spring?
Conservation of Energy
A system of a block of mass 400 𝑔 and a spring with the force constant 50 N/m is oscillating on a frictionless surface as shown in the diagram below. The amplitude of this oscillation is 20 𝑐𝑚. Another block with a mass of 200 𝑔, moving toward the right in the diagram below, strikes the 400 𝑔 mass at the instant when the spring is upstretched and sticks to it on impact. The resulting system is observed to oscillate with an amplitude of 30 𝑐𝑚.
a. What is the period of oscillation when the two blocks stick together?
b. Was the 400 𝑔 mass moving towards the left or right when the other