Question

State the differential equation of linear simple harmonic motion. Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear simple harmonic motion.
A body cools from ${80}^{o}C$ to ${70}^{o}C$ in $5$ minutes and to ${62}^{o}C$ in the next $5$ minutes. Calculate the temperature of the surroundings.

Answer 1

In any SHM, there always exists a restoring force which tries to bring the object back to mean position.

This force causes acceleration in the object.

Hence $F=-kx.$ Here F is the restoring force, x is the displacement of the object from the mean position, and k is the force per unit displacement.

The ${-}_{V}e$ sign indicates that the force is opposite to the displacement. Only then can the object be brought back after displacement.

We know that by Newton's 2nd law of motion, $F=ma$ $\therefore a=\frac{F}{m}$

Applying this, we get $a=-\frac{k}{m}x$

Let $\frac{k}{m}$ be ${\omega }^2$ ------$\omega$ is angular frequency.

Hence we have $a=-{\omega }^{2}x$ expression for acceleration

In calculus $a=\frac{d^{2}x}{d{t}^2}$ ----------the double derivative of displacement with respect to time.

$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$ the differential equation

We can also write $a=\frac{dv}{dt}$---------derivative of velocity w.r.t. time is acceleration.

$\therefore \frac{dv}{dt}=-{\omega }^{2}x$

$\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=v\frac{dv}{dx}$ $\because \frac{dx}{dt}=v$

$\therefore v\frac{dv}{dx}=-{\omega }^{2}x$

$v.dv=-{\omega }^{2}x.dx$

Integrating both sides we get

$\int v.dv=-\int {\omega }^{2}x.dx$

$\frac{v^2}{2}=-\frac{{\omega }^2{x}^2}{2}+C$

In the above equation, when displacement is maximum, x=A and at extreme, v=0

$0=-\frac{{\omega }^2{A}^2}{2}+C$ $\therefore C=\frac{{\omega }^2{A}^2}{2}$

Substituting, we get

$\frac{v^2}{2}=-\frac{{\omega }^2{x}^2}{2}+\frac{{\omega }^2{A}^2}{2}$

$v^2={\omega }^2(A^2-x^2)$

taking root on both sides,

$v=\pm \omega \sqrt{(A^2-x^2)}$ expression for velocity

The $\pm$ indicates that the object can move towards positive direction or negative.

For obtaining the expression for displacement, let us consider only the magnitude of velocity $v=\omega \sqrt{(A^2-x^2)}$

$\frac{dx}{dt}=\omega \sqrt{(A^2-x^2)}$

$\frac{dx}{\sqrt{(A^2-x^2)}}=\omega dt$

integrating both sides, we get

$\int \frac{dx}{\sqrt{(A^2-x^2)}}=\int \omega dt$

$si{n}^{-1}\left(\frac{x}{A}\right)=\omega t+\alpha$ here $\alpha$ is the constant of integration

$\therefore x=Asin(\omega t+\alpha )$ expression for displacement

$\alpha$ is the initial phase angle called as epoch. It depends on initial condition.

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${\theta }_1=\frac{80+70}{2}={75}^{o}C$

${\theta }_2=\frac{70+62}{2}={66}^{o}C$

$\frac{d{\theta }_1}{dt}=\frac{80-70}{5}=2^{o}C/min$

$\frac{d{\theta }_2}{dt}=\frac{70-62}{5}={1.6}^{o}C/min$

$\frac{\left(\frac{d{\theta }_2}{dt}\right)}{\left(\frac{d{\theta }_1}{dt}\right)}=\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}$

$\frac{1.6}{2}=\frac{4}{5}=\frac{66-{\theta }_0}{75-{\theta }_0}$

$(66\times 5)-5{\theta }_0=(75\times 4)-4{\theta }_0$

$(66\times 5)-(75\times 4)={\theta }_0$

$(66\times 5)-(60\times 5)={\theta }_0$

$(66-60)\times 5={30}^{o}C$

ANSWER:- ${30}^{o}C$