Which of the following equations does not represent SHM?
$\text{A.}\quad cos\omega t+ sin \omega t$
$\text{B.}\quad cos\omega t- sin \omega t$
$\text{C.}\quad 1- sin 2\omega t$
$\text{D.}\quad cos(\omega t+\alpha)+ sin \omega t$
Answer
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Hint: A motion in which a particle undergoes periodic motion is called Simple harmonic motion (S.H.M). Not every periodic motion is S.H.M but every S.H.M is periodic motion. Only the motion in which acceleration of a particle is proportional to negative of displacement, that motion is termed as S.H.M (i.e. $a = -\omega^2x$). The revolution of earth about the sun is an example of periodic motion but it is not simple harmonic.
Complete step-by-step answer:
First, let’s understand the standard S.H.M equation.
$Y=asin\left( \omega t+\phi \right)$is called the standard S.H.M equation.
Here ‘Y’ represents the displacement of wave particles at time ‘t’. Coefficient of trigonometric function ‘a’ is called the amplitude of the wave. ‘$\omega$’ is the angular frequency of the wave, which is the measure of angular displacement. ‘$\phi$’ is the initial phase difference of the wave. It is also called ‘epoch’.
Mathematically we can say that if the motion is simple harmonic, it must follow the standard differential equation of simple harmonic motion which is given by $\dfrac{d^2x}{dt^2} = -\omega^2 x$.
If we see for the solution of this equation in higher mathematics, we’ll get the solution of the above equation as $x = Asin (\omega t +\alpha) + B cos(\omega t +\beta)$, based of the values of ‘A’ and ‘B’, we get the equation of S.H.M as of form $Y=asin\left( \omega t+\phi \right)$or $Y=acos\left( \omega t+\phi \right)$.
Now, in options (a), (b) and (d), we can see the equation is in the form of $x = Asin (\omega t +\alpha) + B cos(\omega t +\beta)$.
For (c), we can write
$\quad 1- sin 2\omega t$
$\implies \left( sin^2\omega t + cos^2\omega t\right) - 2sin\omega t cos \omega t$
$\implies \left( sin\omega t - cos\omega t\right)^2$, which is clearly not S.H.M.
Thus, option C. is correct.
So, the correct answer is “Option C”.
Note: Students can make mistakes about the fact that every simple harmonic motion is periodic but not every periodic motion is simple harmonic. The revolution of a fan about its own axis is an example of simple harmonic motion, whereas motion of a pendulum is both periodic and S.H.M.
Complete step-by-step answer:
First, let’s understand the standard S.H.M equation.
$Y=asin\left( \omega t+\phi \right)$is called the standard S.H.M equation.
Here ‘Y’ represents the displacement of wave particles at time ‘t’. Coefficient of trigonometric function ‘a’ is called the amplitude of the wave. ‘$\omega$’ is the angular frequency of the wave, which is the measure of angular displacement. ‘$\phi$’ is the initial phase difference of the wave. It is also called ‘epoch’.
Mathematically we can say that if the motion is simple harmonic, it must follow the standard differential equation of simple harmonic motion which is given by $\dfrac{d^2x}{dt^2} = -\omega^2 x$.
If we see for the solution of this equation in higher mathematics, we’ll get the solution of the above equation as $x = Asin (\omega t +\alpha) + B cos(\omega t +\beta)$, based of the values of ‘A’ and ‘B’, we get the equation of S.H.M as of form $Y=asin\left( \omega t+\phi \right)$or $Y=acos\left( \omega t+\phi \right)$.
Now, in options (a), (b) and (d), we can see the equation is in the form of $x = Asin (\omega t +\alpha) + B cos(\omega t +\beta)$.
For (c), we can write
$\quad 1- sin 2\omega t$
$\implies \left( sin^2\omega t + cos^2\omega t\right) - 2sin\omega t cos \omega t$
$\implies \left( sin\omega t - cos\omega t\right)^2$, which is clearly not S.H.M.
Thus, option C. is correct.
So, the correct answer is “Option C”.
Note: Students can make mistakes about the fact that every simple harmonic motion is periodic but not every periodic motion is simple harmonic. The revolution of a fan about its own axis is an example of simple harmonic motion, whereas motion of a pendulum is both periodic and S.H.M.
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