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Last updated date: 28th Nov 2023
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MVSAT Dec 2023

Show that the motion of a particle represented by\[y=\sin \omega t-\cos \omega t\] is a simple harmonic motion with a time period of \[\dfrac{2\pi }{\omega }\].

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Hint:A simple harmonic motion of an object is an oscillatory motion under a restoring force which is proportional to the displacement of that object from an equilibrium position. We can prove that the given equation represents a S.H.M by comparing the given equation with the standard equation of the S.H.M.

Complete step by step answer:
\[y(t)=Acos(\omega t+\Phi )\] is a standard equation for the simple harmonic motion.
Mathematically the standard equation for SHM is:
\[y(t)=Acos(\omega t+\Phi )\],
where \[y(t)\] is the displacement of the object from its mean position as a function of time, \[A\] is the amplitude or the maximum displacement from the mean position, \[\omega \] is the angular frequency which is equal to \[2\pi f\]and \[f\] is the number of oscillations per second, \[t\] is time in seconds and \[\Phi \] is the phase of the motion or the initial angular displacement of the object from its mean position.

Also \[f=\dfrac{1}{T}\],
where \[T\] is the time period of the motion.
The given equation is \[y=\sin \omega t-\cos \omega t\], taking out \[\sqrt{2}\] from RHS, the equation becomes,
\[y=\sqrt{2}(\dfrac{1}{\sqrt{2}}\sin \omega t-\dfrac{1}{\sqrt{2}}\cos \omega t)\]
\[\Rightarrow y=\sqrt{2}(\cos \dfrac{\pi }{4}\sin \omega t-\sin \dfrac{\pi }{4}\cos \omega t)\]
\[\because \cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}=\sin \dfrac{\pi }{4}\]

Applying the following trigonometric identity to the above equation,
\[\sin (A-B)=\sin A\cos B-\sin B\cos A\]
we get:
\[y=\sqrt{2}(\sin (\omega t-\dfrac{\pi }{4}))\]
Now comparing this equation with the standard equation of SHM,
\[y(t)=Acos(\omega t+\Phi )\]
we can say that the given equation, \[y=\sin \omega t-\cos \omega t\] represents a simple harmonic equation of angular frequency \[\omega \]. Since \[\omega =2\pi f\] and \[f=\dfrac{1}{T}\], where\[T\] is the time period. We can say that the time period of the given equation is \[2\pi /\omega \].

Note:The direction of the restoring force is always towards the equilibrium position, which means that $F=-kx$ where $k$ is a constant of proportion and the negative sign represents the direction of force. Real life examples of SHM include pendulum, swing, spring-mass system, musical instrument etc. If we look around us we can see many other SHMs also.