Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.

Oscillations · Maharashtra HSC

Theory

The differential equation

The differential equation of linear S.H.M. is $\dfrac{d^2x}{dt^2}+\omega^2x=0$

i. Expression for acceleration

a. From the differential equation, $\dfrac{d^2x}{dt^2}=-\omega^2x\qquad\ldots(1)$

b. But the linear acceleration is $\dfrac{d^2x}{dt^2}=a\qquad\ldots(2)$ From equations (1) and (2), $a=-\omega^2x\qquad\ldots(3)$

Equation (3) gives the acceleration in linear S.H.M. Its magnitude is proportional to $x$ and it is always directed towards the mean position.

ii. Expression for velocity

a. From the differential equation of linear S.H.M., $\dfrac{d^2x}{dt^2}=-\omega^2x$ $\therefore \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=-\omega^2x$ $\therefore \dfrac{dv}{dt}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$ $\therefore \dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=-\omega^2x$ $\therefore v\dfrac{dv}{dx}=-\omega^2x$ $\therefore v\,dv=-\omega^2x\,dx\qquad\ldots(4)$

b. Integrating both sides of equation (4), $\int v\,dv=\int-\omega^2x\,dx$ $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+C\qquad\ldots(5)$ where $C$ is the constant of integration.

c. At the extreme position, $x=\pm A$ and $v=0$ . Substituting in equation (5), $0=-\dfrac{\omega^2A^2}{2}+C$ $\therefore C=\dfrac{\omega^2A^2}{2}\qquad\ldots(6)$

d. Substituting equation (6) in equation (5), $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+\dfrac{\omega^2A^2}{2}$ $\therefore v^2=\omega^2(A^2-x^2)$ $\therefore v=\pm\,\omega\sqrt{A^2-x^2}$

This is the required expression for the velocity in linear S.H.M.

iii. Expression for displacement

a. From the above, the velocity is $v=\omega\sqrt{A^2-x^2}\qquad\ldots(1)$ But in linear motion, $v=\dfrac{dx}{dt}\qquad\ldots(2)$ From equations (1) and (2), $\dfrac{dx}{dt}=\omega\sqrt{A^2-x^2}$ $\therefore \dfrac{dx}{\sqrt{A^2-x^2}}=\omega\,dt\qquad\ldots(3)$

b. Integrating both sides of equation (3), $\int\dfrac{dx}{\sqrt{A^2-x^2}}=\int\omega\,dt$ $\therefore \sin^{-1}\left(\dfrac{x}{A}\right)=\omega t+\phi$ where $\phi$ is the constant of integration, which depends on the initial condition (the phase angle at $t=0$ ). $\therefore \dfrac{x}{A}=\sin(\omega t+\phi)$ $\therefore x=A\sin(\omega t+\phi)$

This is the required expression for the displacement of a particle performing linear S.H.M. at time $t$ .

The two starting conditions

Case (i) — the particle starts S.H.M. from the mean position: $x=0$ at $t=0$ , so $\phi=0$ or $\pi$ and $x=\pm A\sin\omega t$ Positive sign if it starts towards the positive side, negative sign if towards the negative side.

Case (ii) — the particle starts S.H.M. from an extreme position: $x=\pm A$ at $t=0$ , so $\phi=\dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$ and $x=\pm A\sin\left(\omega t+\dfrac{\pi}{2}\right)=\pm A\cos\omega t$ Positive sign for the positive extreme position, negative sign for the negative extreme position.

More strictly, it is not the particle that "starts" its S.H.M. — the particle is already performing the motion, and we (the observer) start counting the time from that instant. So $t=0$ is always subjective to the observer.

Expressions at time $t$

From $x=A\sin(\omega t+\phi)$ , $v=\dfrac{dx}{dt}=A\omega\cos(\omega t+\phi)$ $a=\dfrac{dv}{dt}=-A\omega^2\sin(\omega t+\phi)=-\omega^2x$

Extreme values

Displacement: at the mean position $(\omega t+\phi)=0$ or $\pi$ , so $x_{min}=0$ ; at the extreme position $(\omega t+\phi)=\dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$ , so $x_{max}=\pm A$ .
Velocity: at the mean position $x=0$ , so $v_{max}=\pm A\omega$ ; at the extreme positions $x=\pm A$ , so $v_{min}=0$ .
Acceleration: the magnitude is $\omega^2x$ , so at the mean position $a_{min}=0$ and at the extreme positions $a_{max}=\mp A\omega^2$ .

Position	$x$	$v$	$a$
Mean	$0$	$\pm A\omega$ (maximum)	$0$
Extreme	$\pm A$	$0$	$\mp A\omega^2$ (maximum)

Standard results

$a=-\omega^2x$ ; $a_{max}=A\omega^2$
$v=\pm\,\omega\sqrt{A^2-x^2}$ ; $v_{max}=A\omega$
$x=A\sin(\omega t+\phi)$ ; from the mean position $x=\pm A\sin\omega t$ , from an extreme position $x=\pm A\cos\omega t$
$\omega=\dfrac{a_{max}}{v_{max}}$ and $A=\dfrac{v_{max}}{\omega}$
At $v=\dfrac{v_{max}}{2}$ , $x=\pm\dfrac{\sqrt3}{2}A$

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
State the differential equation of linear simple harmonic motion. Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S. H. M.
Maharashtra HSC board question — asked in March 2018Answer 1 ↓
2
From differential equation of linear S.H.M., obtain an expression for acceleration, velocity and displacement of a particle performing S.H.M.
Maharashtra HSC board question — asked in March 2019Answer 2 ↓
3
State the differential equation of linear S.H.M. Hence, obtain expression for : i. acceleration ii. velocity
Maharashtra HSC board question — asked in February 2023Answer 3 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

4
Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by $x=5\sin\left(\dfrac{\pi t}{3}\right)$ m.
Maharashtra HSC board question — asked in February 2023Answer 4 ↓

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.↑ back to the question
Differential equation of linear S.H.M.: The differential equation of linear S.H.M. is given as, $\dfrac{d^2x}{dt^2}+\omega^2x=0$ i. Expression for acceleration in linear S.H.M: a. From differential equation, $\dfrac{d^2x}{dt^2}+\omega^2x=0$ $\dfrac{d^2x}{dt^2}=-\omega^2x\qquad....(1)$ b. But, linear acceleration is given by, $\dfrac{d^2x}{dt^2}=a\qquad....(2)$ From equations (1) and (2), $a=-\omega^2x\qquad....(3)$ Equation (3) gives acceleration in linear S.H.M. ii. Expression for velocity in linear S.H.M: a. From differential equation of linear S.H.M, $\dfrac{d^2x}{dt^2}=-\omega^2x$ $\therefore \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=-\omega^2x$ $\therefore \dfrac{dv}{dt}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$ $\therefore \dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=-\omega^2x$ $\therefore v\dfrac{dv}{dx}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$ $\therefore v\,dv=-\omega^2x\,dx\qquad....(4)$ b. Integrating both sides of equation (4), $\displaystyle\int v\,dv=\int -\omega^2x\,dx$ $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+C\qquad....(5)$ where, C is the constant of integration. c. At extreme position, x = ± A and v = 0. Substituting these values in equation (5), $0=-\dfrac{\omega^2A^2}{2}+C$ $\therefore C=\dfrac{\omega^2A^2}{2}\qquad....(6)$ d. Substituting equation (6) in equation (5), $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+\dfrac{\omega^2A^2}{2}$ $\therefore v^2=\omega^2A^2-\omega^2x^2$ $\therefore v^2=\omega^2(A^2-x^2)$ $\therefore v=\pm\omega\sqrt{A^2-x^2}$ This is the required expression for velocity in linear S.H.M. iii. Expression for displacement in linear S.H.M: a. From differential equation of linear S.H.M, velocity is given by, $v=\omega\sqrt{A^2-x^2}\qquad....(1)$ But, in linear motion, $v=\dfrac{dx}{dt}\qquad....(2)$ From equation (1) and (2), $\dfrac{dx}{dt}=\omega\sqrt{A^2-x^2}$ $\therefore \dfrac{dx}{\sqrt{A^2-x^2}}=\omega\,dt\qquad....(3)$ b. Integrating both sides of equation (3), $\displaystyle\int\dfrac{dx}{\sqrt{A^2-x^2}}=\int\omega\,dt$ $\therefore \sin^{-1}\left(\dfrac{x}{A}\right)=\omega t+\phi$ where, φ is constant of integration which depends upon initial condition (phase angle) $\therefore \dfrac{x}{A}=\sin(\omega t+\phi)$ $\therefore x=A\sin(\omega t+\phi)$ This is required expression for displacement of a particle performing linear S.H.M. at time t.
2.↑ back to the question
[From 5.5, Q1] Differential equation of linear S.H.M.: The differential equation of linear S.H.M. is given as, $\dfrac{d^2x}{dt^2}+\omega^2x=0$ i. Expression for acceleration in linear S.H.M: a. From differential equation, $\dfrac{d^2x}{dt^2}+\omega^2x=0$ $\dfrac{d^2x}{dt^2}=-\omega^2x\qquad....(1)$ b. But, linear acceleration is given by, $\dfrac{d^2x}{dt^2}=a\qquad....(2)$ From equations (1) and (2), $a=-\omega^2x\qquad....(3)$ Equation (3) gives acceleration in linear S.H.M. ii. Expression for velocity in linear S.H.M: a. From differential equation of linear S.H.M, $\dfrac{d^2x}{dt^2}=-\omega^2x$ $\therefore \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=-\omega^2x$ $\therefore \dfrac{dv}{dt}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$ $\therefore \dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=-\omega^2x$ $\therefore v\dfrac{dv}{dx}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$ $\therefore v\,dv=-\omega^2x\,dx\qquad....(4)$ b. Integrating both sides of equation (4), $\displaystyle\int v\,dv=\int -\omega^2x\,dx$ $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+C\qquad....(5)$ where, C is the constant of integration. c. At extreme position, x = ± A and v = 0. Substituting these values in equation (5), $0=-\dfrac{\omega^2A^2}{2}+C$ $\therefore C=\dfrac{\omega^2A^2}{2}\qquad....(6)$ d. Substituting equation (6) in equation (5), $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+\dfrac{\omega^2A^2}{2}$ $\therefore v^2=\omega^2A^2-\omega^2x^2$ $\therefore v^2=\omega^2(A^2-x^2)$ $\therefore v=\pm\omega\sqrt{A^2-x^2}$ This is the required expression for velocity in linear S.H.M. iii. Expression for displacement in linear S.H.M: a. From differential equation of linear S.H.M, velocity is given by, $v=\omega\sqrt{A^2-x^2}\qquad....(1)$ But, in linear motion, $v=\dfrac{dx}{dt}\qquad....(2)$ From equation (1) and (2), $\dfrac{dx}{dt}=\omega\sqrt{A^2-x^2}$ $\therefore \dfrac{dx}{\sqrt{A^2-x^2}}=\omega\,dt\qquad....(3)$ b. Integrating both sides of equation (3), $\displaystyle\int\dfrac{dx}{\sqrt{A^2-x^2}}=\int\omega\,dt$ $\therefore \sin^{-1}\left(\dfrac{x}{A}\right)=\omega t+\phi$ where, φ is constant of integration which depends upon initial condition (phase angle) $\therefore \dfrac{x}{A}=\sin(\omega t+\phi)$ $\therefore x=A\sin(\omega t+\phi)$ This is required expression for displacement of a particle performing linear S.H.M. at time t.
3.↑ back to the question
[From 5.5, Q1 (i)] Differential equation of linear S.H.M.: The differential equation of linear S.H.M. is given as, $\dfrac{d^2x}{dt^2}+\omega^2x=0$ i. Expression for acceleration in linear S.H.M: a. From differential equation, $\dfrac{d^2x}{dt^2}+\omega^2x=0$ $\dfrac{d^2x}{dt^2}=-\omega^2x\qquad....(1)$ b. But, linear acceleration is given by, $\dfrac{d^2x}{dt^2}=a\qquad....(2)$ From equations (1) and (2), $a=-\omega^2x\qquad....(3)$ Equation (3) gives acceleration in linear S.H.M. ii. Expression for velocity in linear S.H.M: a. From differential equation of linear S.H.M, $\dfrac{d^2x}{dt^2}=-\omega^2x$ $\therefore \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=-\omega^2x$ $\therefore \dfrac{dv}{dt}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$ $\therefore \dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=-\omega^2x$ $\therefore v\dfrac{dv}{dx}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$ $\therefore v\,dv=-\omega^2x\,dx\qquad....(4)$ b. Integrating both sides of equation (4), $\displaystyle\int v\,dv=\int -\omega^2x\,dx$ $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+C\qquad....(5)$ where, C is the constant of integration. c. At extreme position, x = ± A and v = 0. Substituting these values in equation (5), $0=-\dfrac{\omega^2A^2}{2}+C$ $\therefore C=\dfrac{\omega^2A^2}{2}\qquad....(6)$ d. Substituting equation (6) in equation (5), $\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+\dfrac{\omega^2A^2}{2}$ $\therefore v^2=\omega^2A^2-\omega^2x^2$ $\therefore v^2=\omega^2(A^2-x^2)$ $\therefore v=\pm\omega\sqrt{A^2-x^2}$ This is the required expression for velocity in linear S.H.M. iii. Expression for displacement in linear S.H.M: a. From differential equation of linear S.H.M, velocity is given by, $v=\omega\sqrt{A^2-x^2}\qquad....(1)$ But, in linear motion, $v=\dfrac{dx}{dt}\qquad....(2)$ From equation (1) and (2), $\dfrac{dx}{dt}=\omega\sqrt{A^2-x^2}$ $\therefore \dfrac{dx}{\sqrt{A^2-x^2}}=\omega\,dt\qquad....(3)$ b. Integrating both sides of equation (3), $\displaystyle\int\dfrac{dx}{\sqrt{A^2-x^2}}=\int\omega\,dt$ $\therefore \sin^{-1}\left(\dfrac{x}{A}\right)=\omega t+\phi$ where, φ is constant of integration which depends upon initial condition (phase angle) $\therefore \dfrac{x}{A}=\sin(\omega t+\phi)$ $\therefore x=A\sin(\omega t+\phi)$ This is required expression for displacement of a particle performing linear S.H.M. at time t. and (ii)
4.↑ back to the question
Given that, $x=5\sin\left(\dfrac{\pi t}{3}\right)$ m Velocity of a particle performing S.H.M. is given by, $v=\dfrac{dx}{dt}=5\times\dfrac{d}{dt}\left[\sin\left(\dfrac{\pi t}{3}\right)\right]=5\times\cos\left(\dfrac{\pi t}{3}\right)\times\dfrac{\pi}{3}$ In t = 1 second, $v=5\times\dfrac{1}{2}\times\dfrac{\pi}{3}=2.62$ m/s