Two simple harmonic motions are represented by the equations \${Y_1} = 10\;\sin \left( {3\pi t + \dfrac{\pi }{4}} \right)\$ and \${Y_2} = 5\left( {\sin 3\pi t + \sqrt 3 \cos \;3\pi t} \right)\$Their amplitudes are in the ratio of;

Given, \$Y_1=10sin(3\pi t+\pi/4)\$ and \$Y_2=5(sin3\pi t+\sqrt3 cos3\pi t)=5sin3\pi t+5\sqrt3 cos3\pi t\$Amplitude of \$Y_1\$ is \$A_1=10\$ and amplitude of \$Y_2\$ is \$A_2=\sqrt{(5)^2+(5\sqrt 3)^2}=\sqrt{25+75}=10\$Thus, \$\dfrac{A_1}{A_2}=\dfrac{10}{10}=1:1\$

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>>Class 11
>>Physics
>>Oscillations
>>Energy in SHM
>> $Graphical variation of Energy in SHM$

Graphical variation of Energy in SHM

Q: Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

For SHM, acceleration , a = -ω²y F = ma = -mω²y now, work , W = F.dy cos180° { because displacement and acceleration are in opposite direction so, cos180° taken } W = ∫mω²y.dy = mω²y²/2 use standard form of SHM , y = Asin(ωt ± Ф) W = mω²A²/2 sin²(ωt ± Ф) We know, Potential energy is work done stored in system . so, P.E = W = mω²A²/2 sin²(ωt ± Ф) again, Kinetic energy , K.E = 1/2mv² , here v is velocity we know, v = ωAcos(ωt ± Ф) so, K.E = mω²A²/2cos²(ωt ± Ф) Total energy = K.E + P.E = mω²A²/2 cos²(ωt ± Ф) + mω²A²/2 sin²(ωt ± Ф) = mω²A²/2 [ cos²(ωt ± Ф) + sin²(ωt ± Ф) ] = mω²A²/2 [ ∵sin²α + cos²α = 1] = mω²A²/2 = constant

Q: In a SHM, when the displacement is one half the amplitude, what fraction of the total energy is kinetic?

In case of SHM\$\displaystyle KE = \frac{1}{2} m \omega^2 (a^2 - x^2)\$and at \$x =a/2\$\$\displaystyle KE = \frac{1}{2} m \omega^2 [a^2 - (a/2)^2]\$\$\displaystyle = \frac{3}{4} \left[ \frac{1}{2} m \omega^2 a^2 \right]\$Total energy = \$\displaystyle \frac{1}{2} m \omega^2 a^2\$\$\therefore\$ Fraction of total energy\$\displaystyle = \frac{KE}{Total \,\, energy} = \frac{\displaystyle \frac{3}{4} \left[ \frac{1}{2}m \omega^2 a^2 \right] }{\displaystyle \frac{1}{2}m \omega a^2} = \frac{3}{4}\$

Q: The kinetic energy and potential energy of a particle executing SHM will be equal when displacement (\$amplitude =a\$) is

\$K.E.=P.E.\\ \Rightarrow \frac { 1 }{ 2 } m{ \omega }^{ 2 }({ a }^{ 2 }-{ x }^{ 2 })=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ x }^{ 2 }\\ \Rightarrow { a }^{ 2 }-{ x }^{ 2 }={ x }^{ 2 }\\ \Rightarrow 2{ x }^{ 2 }={ a }^{ 2 }\\ \Rightarrow x=a/\sqrt { 2 } \$

Q: Two springs of spring constant \$1500\ N/m\$ and \$3000\ N/m\$ respectively are stretched with the same force. They will have the potential energies in the ratio of

\$KE=\dfrac{1}{2}*k*x^2\$\$k_1=3000N/m,k_2=1500N/m\$\$KE_1:KE_2=k_1:k_2=3000:1500=2:1\$ for same \$x\$

Q: Starting from an origin, a body oscillates simple harmonically with a period of \$2s\$. After what time will its kinetic energy be \$75\%\$ of the total energy?

As per, \$ \displaystyle KE = \frac {75}{100} E = \frac {75}{100} \times \frac {1}{2} ma^2 \omega^2 \$\$\displaystyle \frac {1}{2} ma^2 \omega^2 \cos^2 \omega t = \frac {75}{100} \times \frac {1}{2} ma^2 \omega^2 \$\$ \cos^2 \omega t = \dfrac {3}{4} \$\$ \cos \omega t = \dfrac {\sqrt3}{2} = \cos \dfrac {\pi}{6} \$or \$ \omega t = \dfrac {\pi}{6} \Rightarrow t = \dfrac {\pi}{6 \pi} = \dfrac {1}{6} s \$

Q: A block whose mass is \$1kg\$ is fastened to a spring. The spring has a spring constant of \$50{Nm}^{-1}\$. The block is pulled to a distance of \$x=10cm\$ from its equilibrium position at \$x=0\$ \$cm\$ on a frictionless surface from rest at \$t=0\$. The kinetic energy of the block when it is \$5cm\$ away from the mean position is

Angular frequency of S.H.M\$=\sqrt { \cfrac { k }{ { m } } } =5\sqrt { 2 } \$Here, the body is displaced to maximum at \$x=10cm=0.1m\$. So, it is the amplitude of the motion velocity of the body performing S.H.M at any position \$x\$ then mean position is \$v=\omega \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \$Here \$a=0.1m\$\$x=5cm=0.05m\$\$\Rightarrow v=5\sqrt { 2 } (\sqrt { 0.01-0.0005 } )\$\$v=0.61m/s\$Kinetic energy at this point\$=\cfrac { 1 }{ 2 } m{ v }^{ 2 }=\cfrac { 1 }{ 2 } \times 1\times { (0.61) }^{ 2 }\simeq 0.19J\$

Q: Obtain expressions for kinetic energy, potential energy and total energy of a particle performing linear S.H.M.

Acceleration of the particle , performing S.H.M is given by \$\alpha = -\omega^2 y\$ where \$\omega\$ is the angular velocity, and y is the displacement of particle. now, workdone by particle = \$\vec F. \vec d y\$ as we know, acceleration and displacement are in opposite directions in case of S.H.M so, \$W = -m\omega^2 y dy\$ where m is the mass of the particle. \$W = -m\omega^2 \int ydy\$ \$W = - \dfrac{1}{2} m \omega^2 y^2\$ so, potential energy = -W \$=\dfrac{1}{2} m\omega^2 y^2\$ we know, \$\omega = 2 \pi \eta\$ so, \$P.E = 2 \pi^2 \eta^2 m y^2\$ ......(1) velocity of particle , \$v = \omega Acos \omega t\$ or, \$v = \omega \sqrt{A^2 - y^2}\$ so, kinetic energy of particle, \$K.E = \dfrac{1}{2} mv^2\$ hence, \$K.E = \dfrac{1}{2} m \omega^2 ( A^2 - y^2)\$ but \$\omega = 2 \pi \eta\$ so, \$K.E = 2 \pi^2 \eta^2 m ( A^2 - y^2)\$ ....(2) so, total mechanical energy = K.E + P.E \$= 2 \pi^2 \eta^2 mA^2\$ ......(3)

Q: The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5J. If its total energy is 9J and its amplitude is 0.01 m,its time period would be

kinetic energy at mean position= total energy-potential energy at mean position\$=9J-5J=4J\$kinetic energy at mean position\$=\dfrac { 1 }{ 2 } m{ v_{ max } }^{ 2 }\$\$\Rightarrow \dfrac { 1 }{ 2 } m{ v_{ max } }^{ 2 }=4J\$\$\Rightarrow { v_{ max } }=\sqrt { \dfrac { 4\times 2 }{ 2 } } m/s=2m/s\$\$\Rightarrow A\omega =A\dfrac { 2\pi }{ T } \\ \Rightarrow A\dfrac { 2\pi }{ T } =2m/s\\ \Rightarrow T=A\pi =0.01\pi =\dfrac { \pi }{ 100 } s\$

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The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5J. If its total energy is 9J and its amplitude is 0.01 m,its time period would be

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Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

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

In a SHM, when the displacement is one half the amplitude, what fraction of the total energy is kinetic?

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement (amplitude=a) is

Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have the potential energies in the ratio of

Starting from an origin, a body oscillates simple harmonically with a period of 2s. After what time will its kinetic energy be 75% of the total energy?

Two simple harmonic motions are represented by the equations Y1​=10sin(3πt+4π​) and Y2​=5(sin3πt+3​cos3πt) Their amplitudes are in the ratio of;

Obtain expressions for kinetic energy, potential energy and total energy of a particle performing linear S.H.M.

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5J. If its total energy is 9J and its amplitude is 0.01 m,its time period would be

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement ( $a m p l i t u d e = a$ ) is

Two springs of spring constant $1500 N / m$ and $3000 N / m$ respectively are stretched with the same force. They will have the potential energies in the ratio of

Starting from an origin, a body oscillates simple harmonically with a period of $2 s$ . After what time will its kinetic energy be $75 %$ of the total energy?

Two simple harmonic motions are represented by the equations
$Y_{1} = 10 sin (3 π t + \frac{π}{4})$ and $Y_{2} = 5 (sin 3 π t + 3 cos 3 π t)$
Their amplitudes are in the ratio of;