Energy of a Particle Performing S.H.M.

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   The question is a real board question. Its answer and working are ours — this paper was published without an official answer key.

   The Three Energies

   For a particle of mass $m$ performing linear S.H.M. of amplitude $A$ and angular frequency $\omega$, with force constant $k = m\omega^2$, at a displacement $x$ from the mean position:

   $$E_k = \dfrac{1}{2}k\left(A^2 - x^2\right), \qquad E_p = \dfrac{1}{2}kx^2, \qquad E = E_k + E_p = \dfrac{1}{2}kA^2$$

   The total energy is independent of $x$, so it is conserved throughout the motion.

   The Graph

   • Axes: the instantaneous displacement $x$ along the horizontal axis, from $-A$ on the left, through the mean position $O$ at $x = 0$, to $+A$ on the right; energy along the vertical axis.
   • P.E. curve: $E_p = \dfrac{1}{2}kx^2$ is an upward-opening parabola with its vertex (value zero) at $x = 0$, rising symmetrically to $\dfrac{1}{2}kA^2$ at $x = -A$ and at $x = +A$.
   • K.E. curve: $E_k = \dfrac{1}{2}k\left(A^2 - x^2\right)$ is a downward-opening parabola with its maximum $\dfrac{1}{2}kA^2$ at $x = 0$, falling symmetrically to zero at $x = \pm A$. It is the mirror image of the P.E. curve about the horizontal line at $\dfrac{1}{4}kA^2$.
   • T.E. line: $E = \dfrac{1}{2}kA^2$ is a horizontal straight line at that height, the same at every value of $x$; the two parabolas add up to it everywhere.
   • The two parabolas intersect at $x = \pm\dfrac{A}{\sqrt2}$, where $E_k = E_p = \dfrac{E}{2} = \dfrac{1}{4}kA^2$. Label these crossing points.
   • Label all three curves K.E., P.E. and T.E., and mark $-A$, $O$ and $+A$ on the displacement axis and $\dfrac{1}{2}kA^2$ on the energy axis.

   Values at the Key Points

   Position	$x$	K.E.	P.E.	T.E.
   Extreme (left)	$-A$	$0$	$\dfrac{1}{2}kA^2$	$\dfrac{1}{2}kA^2$
   Crossing point	$\pm\dfrac{A}{\sqrt2}$	$\dfrac{1}{4}kA^2$	$\dfrac{1}{4}kA^2$	$\dfrac{1}{2}kA^2$
   Mean	$0$	$\dfrac{1}{2}kA^2$	$0$	$\dfrac{1}{2}kA^2$
   Extreme (right)	$+A$	$0$	$\dfrac{1}{2}kA^2$	$\dfrac{1}{2}kA^2$

   Conclusion

   The energy is wholly kinetic at the mean position and wholly potential at the two extreme positions; between them it is continually converted from one form to the other, but their sum, the total energy, stays constant. Both parabolas are symmetric about the energy axis, so the energies depend only on the magnitude of the displacement, not on its direction.

2. 2.↑ back to the question

   Refer Subtopic 5.11: Q. No. 1 for graphical representation. i. Kinetic energy of particle executing S.H.M is given by, $E_k=\dfrac{1}{2}k(A^2-x^2)=\dfrac{1}{2}m\omega^2(A^2-x^2)$ ii. Potential energy of particle executing S.H.M. is given by, $E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$ iii. For a particle performing linear S.H.M. If $E_k=E_p$, $\dfrac{1}{2}m\omega^2(A^2-x^2)=\dfrac{1}{2}m\omega^2x^2$ $\therefore x=\pm\dfrac{A}{\sqrt2}$ Thus at $x=\pm\dfrac{A}{\sqrt2}$, $E_k=E_p=\dfrac{E}{2}$

3. 3.↑ back to the question

   Expression for potential energy: i. The restoring force acting on the particle at point P is given by, $f=-kx$ where k is the force constant. ii. Suppose that the particle is displaced further by an infinitesimal displacement 'dx' against the restoring force 'f'. iii. The external work done (dW) during this displacement is $dW=f(-dx)=-kx(-dx)=kx\,dx$ iv. The total work done on the particle to displace it from O to P is given by, $$W=\int_0^x dW=\int_0^x kx\,dx=\dfrac{1}{2}kx^2$$ v. This work done is stored as the potential energy (P.E.) $E_p$ of the particle at displacement x. $\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$ vi. At time t, $\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}kA^2\sin^2(\omega t+\phi)$ $=\dfrac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$ Thus, with time, it varies as $\sin^2\theta$. vii. At the mean position, x = 0 $\therefore$ Potential energy, $(E_p)_{min}=0$ viii. At the extreme positions, x = ± A $\therefore$ Potential energy, $(E_p)_{max}=\dfrac{1}{2}m\omega^2A^2$

4. 4.↑ back to the question

   i. Expression for kinetic energy: a. Consider a particle of mass m, performing a linear S.H.M. along the path MN about the mean position O as shown in figure. b. At a given instant, let the particle be at P, at a distance x from O. c. Velocity of the particle in S.H.M. is given as $v=\omega\sqrt{A^2-x^2}=A\omega\cos(\omega t+\phi)$ where x is the displacement of the particle performing S.H.M. and A is the amplitude of S.H.M. d. Thus, the kinetic energy, $E_k=\dfrac{1}{2}mv^2$ $=\dfrac{1}{2}m\omega^2(A^2-x^2)\qquad....(1)$ $=\dfrac{1}{2}k(A^2-x^2)$ This is the kinetic energy at displacement x. e. Also, at time t, kinetic energy is, $E_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$ $=\dfrac{1}{2}kA^2\cos^2(\omega t+\phi)$ Thus, with time, it varies as $\cos^2\theta$. ii. Expression for potential energy: [From 5.11, Q3] Expression for potential energy: i. The restoring force acting on the particle at point P is given by, $f=-kx$ where k is the force constant. ii. Suppose that the particle is displaced further by an infinitesimal displacement 'dx' against the restoring force 'f'. iii. The external work done (dW) during this displacement is $dW=f(-dx)=-kx(-dx)=kx\,dx$ iv. The total work done on the particle to displace it from O to P is given by, $$W=\int_0^x dW=\int_0^x kx\,dx=\dfrac{1}{2}kx^2$$ v. This work done is stored as the potential energy (P.E.) $E_p$ of the particle at displacement x. $\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$ vi. At time t, $\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}kA^2\sin^2(\omega t+\phi)$ $=\dfrac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$ Thus, with time, it varies as $\sin^2\theta$. vii. At the mean position, x = 0 $\therefore$ Potential energy, $(E_p)_{min}=0$ viii. At the extreme positions, x = ± A $\therefore$ Potential energy, $(E_p)_{max}=\dfrac{1}{2}m\omega^2A^2$[(i) to (vi)] iii. Expression for total energy: a. The total energy of the particle is the sum of its kinetic energy and potential energy. $\therefore E=E_k+E_p$ b. Using equation (1) and equation (2), we get $E=\dfrac{1}{2}m\omega^2(A^2-x^2)+\dfrac{1}{2}m\omega^2x^2$ $E=\dfrac{1}{2}m\omega^2A^2=\dfrac{1}{2}kA^2=\dfrac{1}{2}m(v_{max})^2\qquad....(3)$ This expression gives the total energy of the particle at point P. iv. As m, ω and A are constant, the total energy of the particle at any point P is constant (independent of x and t). In other words, the energy is conserved in S.H.M. Graphical representation of K.E. and P.E. of S.H.M.: Refer Subtopic 5.11: Q. No. 1

5. 5.↑ back to the question

   [From 5.11, Q3] Expression for potential energy: i. The restoring force acting on the particle at point P is given by, $f=-kx$ where k is the force constant. ii. Suppose that the particle is displaced further by an infinitesimal displacement 'dx' against the restoring force 'f'. iii. The external work done (dW) during this displacement is $dW=f(-dx)=-kx(-dx)=kx\,dx$ iv. The total work done on the particle to displace it from O to P is given by, $$W=\int_0^x dW=\int_0^x kx\,dx=\dfrac{1}{2}kx^2$$ v. This work done is stored as the potential energy (P.E.) $E_p$ of the particle at displacement x. $\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$ vi. At time t, $\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}kA^2\sin^2(\omega t+\phi)$ $=\dfrac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$ Thus, with time, it varies as $\sin^2\theta$. vii. At the mean position, x = 0 $\therefore$ Potential energy, $(E_p)_{min}=0$ viii. At the extreme positions, x = ± A $\therefore$ Potential energy, $(E_p)_{max}=\dfrac{1}{2}m\omega^2A^2$

6. 6.↑ back to the question

   Definition: The physical quantity which describes the state of oscillation of a particle performing S.H.M at the start of motion is called epoch of S.H.M. Factors affecting total energy: Total energy of particle performing S.H.M. depends on the mass, amplitude and frequency of motion of the particle.

7. 7.↑ back to the question

   At the extreme position, the total energy of a particle executing linear S.H.M. is purely potential.

8. 8.↑ back to the question

   Given: T = 10 s, $P.E=\dfrac{1}{2}(T.E)$ To find: Time (t) Formulae: i. $P.E=\dfrac{1}{2}m\omega^2x^2$ ii. $T.E=\dfrac{1}{2}m\omega^2A^2$ Calculation: From formula (i) and (ii), $\dfrac{1}{2}m\omega^2x^2=\dfrac{1}{2}\left(\dfrac{1}{2}m\omega^2A^2\right)$ ... (From given condition) $\therefore x=\dfrac{A}{\sqrt2}$ As, $x=A\sin\omega t$ $A\sin\omega t=\dfrac{A}{\sqrt2}$ $\therefore \sin\omega t=\dfrac{1}{\sqrt2}$ $\therefore \omega t=\sin^{-1}\left(\dfrac{1}{\sqrt2}\right)$ $\therefore \left(\dfrac{2\pi}{T}\right)t=\dfrac{\pi}{4}\qquad\left(\because\omega=\dfrac{2\pi}{T}\right)$ $\therefore t=\dfrac{T}{8}=\dfrac{10}{8}$ $\therefore t=1.25$ s Ans: The time in which the potential energy will be half of total energy is 1.25 s.

9. 9.↑ back to the question

   Given: x = A/3 To find: Ratio of kinetic and total energy $\left(\dfrac{K.E.}{T.E.}\right)$ Formulae: i. $T.E.=\dfrac{1}{2}kA^2$ ii. $K.E.=\dfrac{1}{2}k(A^2-x^2)$ Calculation: From formula (ii), $K.E.=\dfrac{1}{2}k\left[A^2-\left(\dfrac{A}{3}\right)^2\right]=\dfrac{1}{2}k\left[A^2-\dfrac{A^2}{9}\right]$ $=\dfrac{1}{2}k\times\dfrac{8A^2}{9}$ From formulae (i) and (ii), $\dfrac{K.E.}{T.E.}=\dfrac{\dfrac{1}{2}kA^2\left(\dfrac{8}{9}\right)}{\dfrac{1}{2}kA^2}=\dfrac{8}{9}$ Ans: The ratio of kinetic and total energy is 8 : 9.