A function of time given by $\left( {\sin \omega t - \cos \omega t} \right)$ represents:
1) Simple Harmonic Motion.
2) Non-periodic Motion.
3) Periodic but not simple harmonic motion
4) Oscillatory but not simple harmonic motion
Answer
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Hint: In this question one has to know to compare the given equation $\left( {\sin \omega t - \cos \omega t} \right)$ with the general equation of Simple harmonic motion. If the equation matches with the general equation of Simple Harmonic Motion then the given equation is of SHM otherwise it is not.
Complete step by step solution:
The general form of SHM is given by:
$y = A\sin (\omega t + \phi )$;
We have been given the equation as:
y =$\left( {\sin \omega t - \cos \omega t} \right)$;
Multiply the above equation on the RHS with$\sqrt 2 $.
$y = \sqrt 2 \left( {\dfrac{{\sin \omega t}}{{\sqrt 2 }} - \dfrac{{\cos \omega t}}{{\sqrt 2 }}} \right)$;
Here the value$\dfrac{1}{{\sqrt 2 }}$is equal to$\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4}$; Put this value in the above equation:
$y = \sqrt 2 \left( {\sin \omega t \times \cos \dfrac{\pi }{4} - \cos \omega t \times \sin \dfrac{\pi }{4}} \right)$;
Apply trigonometric property $(\sin a\cos b - \cos b\sin a) = \sin (a - b)$in the above equation:
$y = \sqrt 2 \left( {\sin (\omega t - \dfrac{\pi }{4})} \right)$;
The above equation is similar to the general equation for SHM which is:
$y = A\sin (\omega t + \phi )$;
Compare the above two equations with each other.
Here: - $A = \sqrt 2 $;$\phi = \dfrac{\pi }{4}$
Hence the option 2,3 and 4 are ruled out as they are not depicting Simple Harmonic Motion.
After comparing the above two equations we came to the conclusion that the equation $\left( {\sin \omega t - \cos \omega t} \right)$represents Simple Harmonic Motion with a time period$\dfrac{{2\pi }}{\omega }$.
Final Answer:Option “1” is correct. A function of time given by $\left( {\sin \omega t - \cos \omega t} \right)$represents Simple Harmonic Motion.
Note:In Simple Harmonic Motion there is a force which tries to bring back an object to its mean position this force is known as restoring force. In SHM the restoring force is proportional to the displacement of the object from its original or mean position. The direction of the force that is restoring in nature is always directed towards the mean position.
Complete step by step solution:
The general form of SHM is given by:
$y = A\sin (\omega t + \phi )$;
We have been given the equation as:
y =$\left( {\sin \omega t - \cos \omega t} \right)$;
Multiply the above equation on the RHS with$\sqrt 2 $.
$y = \sqrt 2 \left( {\dfrac{{\sin \omega t}}{{\sqrt 2 }} - \dfrac{{\cos \omega t}}{{\sqrt 2 }}} \right)$;
Here the value$\dfrac{1}{{\sqrt 2 }}$is equal to$\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4}$; Put this value in the above equation:
$y = \sqrt 2 \left( {\sin \omega t \times \cos \dfrac{\pi }{4} - \cos \omega t \times \sin \dfrac{\pi }{4}} \right)$;
Apply trigonometric property $(\sin a\cos b - \cos b\sin a) = \sin (a - b)$in the above equation:
$y = \sqrt 2 \left( {\sin (\omega t - \dfrac{\pi }{4})} \right)$;
The above equation is similar to the general equation for SHM which is:
$y = A\sin (\omega t + \phi )$;
Compare the above two equations with each other.
Here: - $A = \sqrt 2 $;$\phi = \dfrac{\pi }{4}$
Hence the option 2,3 and 4 are ruled out as they are not depicting Simple Harmonic Motion.
After comparing the above two equations we came to the conclusion that the equation $\left( {\sin \omega t - \cos \omega t} \right)$represents Simple Harmonic Motion with a time period$\dfrac{{2\pi }}{\omega }$.
Final Answer:Option “1” is correct. A function of time given by $\left( {\sin \omega t - \cos \omega t} \right)$represents Simple Harmonic Motion.
Note:In Simple Harmonic Motion there is a force which tries to bring back an object to its mean position this force is known as restoring force. In SHM the restoring force is proportional to the displacement of the object from its original or mean position. The direction of the force that is restoring in nature is always directed towards the mean position.
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