Simple Harmonic Motion Concept Page - 5

Example

Derive and use kinetic energy per unit length at a given point in a travelling sound wave

Example:
The velocity of a sound wave in v${\displaystyle v}$ and the wave energy density is E,${\displaystyle E,}$ then find the amount of energy transferred per unit area per second by the wave in a direction normal to the wave propagation.

Solution:
Energy transferred in normal direction to the wave propogation is also known as intensity 
I=12ρa2ω2v$I={\displaystyle \frac{1}{2}}\rho a^2{\omega }^{2}v$
E=ω2a2ρ$E={\omega }^2{a}^2\rho$
hence I=Ev$I=Ev$

Formula

Total Potential Energy in one wavelength in a travelling sound wave

Total potential energy of a traveling sound wave is given by:
P.E.=14μω2A2λ$P.E.={\displaystyle \frac{1}{4}}\mu {\omega }^2{A}^2\lambda$
 where,
 μ=$\mu =$ Linear mass density,
ω=$\omega =$ Angular frequency,
A=$A=$ Amplitude of wave.

Example

Kinetic energy per unit length at a given point in a travelling sound wave

Kinetic Energy of a traveling sound wave is defined as:
K.E.=∫ρv22dV$K.E.=\int {\displaystyle \frac{\rho v^2}{2}}dV$

Example. Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in figure. The speed of each pulse is 2 cm/s. After 2 seconds, what will be the total energy of the pulses.

Solution:After two seconds, the two pulses would nullify each other. As the string now becomes straight, there would be no deformation of the string. In such a situation, there would be no potential energy.

Definition

Average kinetic energy and potential energy of a SHM

Total energy in SHM is given by, E=12mω2A2$E={\displaystyle \frac{1}{2}}m{\omega }^2{A}^2$ where A$A$ is the amplitude and remains conserved.
E=K+U$E=K+U$
Kavg=Uavg=E2=14mω2A2$K_{avg}=U_{avg}={\displaystyle \frac{E}{2}}={\displaystyle \frac{1}{4}}m{\omega }^2{A}^2$
Note:
Average kinetic energy can also be found using Kavg=1T∫T0Kdt$K_{avg}={\displaystyle \frac{1}{T}}{\int }_0^{T}Kdt$
Average potential energy can also be found using Uavg=1T∫T0Udt$U_{avg}={\displaystyle \frac{1}{T}}{\int }_0^{T}Udt$

Diagram

Plot kinetic energy, potential energy and total energy against displacement

Example

Write kinetic energy as a function of time in SHM

Kinetic energy as a function of time in SHM:

E = mw2(A2−A2sin2(wt+ϕ))2${\displaystyle \frac{m{w}^2(A^2-A^{2}si{n}^2(wt+\varphi ))}{2}}$

Formula

Write potential energy as a function of time in SHM

Formula for potential energy as a function of time in SHM is:

P.E.=mw2x2sin2(wt+ϕ)2$P.E.={\displaystyle \frac{m{w}^2{x}^{2}si{n}^2(wt+\varphi )}{2}}$

Formula

Total energy as a function of time in SHM

Total energy as a function of time in SHM:

Total energy = mw2(A2)2${\displaystyle \frac{m{w}^2(A^2)}{2}}$ (Independent of time)

Example

Conservation of total mechanical energy to find amplitude

Example: Potential energy of a particle in SHM along x−$x-$axis is given by: U=10+(x−2)2$U=10+{(x-2)}^2$ .Here, U$U$ is in joule and x$x$ in meter. Total mechanical energy of the particle is 26 J.$26J.$ Mass of the particle is 2 kg$2kg$ Find the amplitude of oscillation.

Solution:
U0=$U_0=$minimum potential energy at mean position (x=2)=10 J$(x=2)=10J$
At extreme position
U=$U=$ Total mechanical energy =26 J$=26J$
 =10+(x−2)2$=10+{(x-2)}^2$
∴$\therefore$   (x−2)=±4$(x-2)=\pm 4$
Hence x=6 m$x=6m$ and x=−2 m$x=-2m$ are the extreme positions.
Amplitude of oscillation=4 m$=4m$

Example

Problem on kinetic energy, potential energy and total energy of a mass attached to a spring in SHM

Example: A mass m$m$ is attached to a spring of stiffness k$k$ executing SHM. It has amplitude A$A$ and velocity at the equilibrium position is u$u$. Find the total energy of this spring mass system.

Solution:
At the extreme position of the spring it has only potential energy since velocity is zero: kA22${\displaystyle \frac{k{A}^2}{2}}$
At the equilibrium position it has no stretch in the spring.
Kinetic energy at this instant: mu22${\displaystyle \frac{m{u}^2}{2}}$
At any instant of time during the motion:
Total energy = KE + PE = mu22${\displaystyle \frac{m{u}^2}{2}}$ = kA22${\displaystyle \frac{k{A}^2}{2}}$

Example

Write kinetic energy, potential energy and total energy of a mass attached to a spring in SHM

Kinetic Energy in SHM: mw2(A2−x2)2${\displaystyle \frac{m{w}^2(A^2-x^2)}{2}}$
Potential Energy is : mw2(x2)2${\displaystyle \frac{m{w}^2(x^2)}{2}}$
Total Energy is: mw2(A2)2${\displaystyle \frac{m{w}^2(A^2)}{2}}$

Example: The potential energy of a simple pendulum in its resting position is 10$10$ J and its mean kinetic energy is 5$5$ J. What will be its total energy at any instant?

Solution:
The total energy of the system remains constant. Since it is given that P.E at rest is 10$10$ J, the total energy must be 10$10$ J as K.E at rest is 0. As total energy of the system is conserved in SHM.

Definition

Potential energy of a spring

A spring stores potential energy due to extension. Since an unextended spring does not store potential energy, it is used as the point of zero energy. 
For a spring, potential energy is defined as U=12kx2$U={\displaystyle \frac{1}{2}}k{x}^2$ where x$x$ is the extension of the spring.

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