Question

If the displacement of a moving particle at any time be given by an equation $x=a\cos \omega t+b\sin \omega t$, show that the motion is simple harmonic. If $a=3,b=4,\omega =2$, determine the time period, amplitude, maximum velocity and maximum acceleration.

Answer 1

Hint:Differentiate the given equation of displacement with respect to $x$ two times, to find the acceleration of the particle. Then compare the equation of acceleration with the equation of acceleration of a particle in SHM.

Formula used:
$a={\displaystyle \frac{d^{2}x}{d{t}^2}}$
Here $a$ is acceleration, $x$ is displacement and $t$ is time.

Complete step by step answer:
A particle is said to be in a simple harmonic motion when it acceleration can be written as
$a=-{\omega }^{2}x$ ….. (i),
where $a$ is the acceleration of the particle, $x$ is its displacement and $\omega$ is a proportionality constant called the angular frequency of the motion.
The acceleration of a particle, in mechanics, is given as $a={\displaystyle \frac{d^{2}x}{d{t}^2}}$.
Therefore, let us find the double derivative of the given equation for the displacement of the particle and check whether the result is similar to equation (i).The displacement of the particle is given to be $x=a\cos \omega t+b\sin \omega t$ ….. (ii).

Now, differentiate (ii) with respect to x. Then,
${\displaystyle \frac{dx}{dt}}={\displaystyle \frac{d}{dt}}(a\cos \omega t+b\sin \omega t)$
$\Rightarrow {\displaystyle \frac{dx}{dt}}=-a\omega \sin \omega t+b\omega \cos \omega t$
Now, again differentiate the above equation with respect to x.
With this we get,
${\displaystyle \frac{d^{2}x}{d{t}^2}}={\displaystyle \frac{d}{dt}}(-a\omega \sin \omega t+b\omega \cos \omega t)$
$\Rightarrow {\displaystyle \frac{d^{2}x}{d{t}^2}}=-a{\omega }^2\cos \omega t-b{\omega }^2\sin \omega t$
Then,
$\Rightarrow acceleration=-a{\omega }^2\cos \omega t-b{\omega }^2\sin \omega t$
$\Rightarrow acceleration=-{\omega }^2(a\cos \omega t+b\sin \omega t)$
But, we know that $x=a\cos \omega t+b\sin \omega t$.
Then,
$\Rightarrow acceleration=-{\omega }^{2}x$
The above equation is the equation for acceleration of a particle undergoing simple harmonic motion.
Therefore, the particle is undergoing simple harmonic motion.
When the equation of displacement of the particle in simple harmonic motion is in the form $x=a\cos \omega t+b\sin \omega t$, the amplitude of the motion is given as $A=\sqrt{a^2+b^2}$.
Therefore, the amplitude of the given motion is $A=\sqrt{3^2+4^2}=\sqrt{25}=5$
The time period of SHM is given as $T={\displaystyle \frac{2\pi }{\omega }}$.
Therefore, the time period in this case is $T={\displaystyle \frac{2\pi }{2}}=\pi$
Maximum velocity in SHM is given as $v_{max}=A\omega$.
Therefore, the maximum velocity in this case is $v_{max}=(5)(2)=10$
Maximum acceleration in SHM is given as $a_{max}=A{\omega }^2$.

Therefore, the maximum acceleration in this case is $v_{max}=(5){(2)}^2=20$.

Note: The general equation for the displacement of the particle in SHM is given as $x=A\sin (\omega t+\varphi )$, where A is amplitude, $\omega$ is angular frequency and $\varphi$ is a constant called phase constant. We can divide and multiply the right hand side of equation (i) by $\sqrt{a^2+b^2}$
Then,
$\Rightarrow x=\sqrt{a^2+b^2}({\displaystyle \frac{a}{\sqrt{a^2+b^2}}}\cos \omega t+{\displaystyle \frac{b}{\sqrt{a^2+b^2}}}\sin \omega t)$ …. (iii)
Now, we can write ${\displaystyle \frac{a}{\sqrt{a^2+b^2}}}=\sin \varphi$ and ${\displaystyle \frac{b}{\sqrt{a^2+b^2}}}=\cos \varphi$, since a and b are constants.
Substitute these values in (iii)
$\Rightarrow x=\sqrt{a^2+b^2}(\sin \varphi \cos \omega t+\cos \varphi \sin \omega t)$
We know that's $(\sin \varphi \cos \omega t+\cos \varphi \sin \omega t)=\sin (\omega t+\varphi )$
Then,
$\Rightarrow x=\sqrt{a^2+b^2}\sin (\omega t+\varphi )$
Now, we can write $\sqrt{a^2+b^2}=A$
Then,
$\Rightarrow x=A\sin (\omega t+\varphi )$
This is another way in which we can show that the particle executes SHM.