Question

Two SHMs of the same period and amplitudes ${a_1}$ and ${a_2}$ act on a particle simultaneously parallel to each other. The resultant motion is
A. SHM of double period
B. SHM of same period
C. Vibratory motion but not SHM
D. A periodic motion

Answer 1

Hint: In order to solve this question we need to understand Simple Harmonic motion and superposition principle. Simple Harmonic Motion is periodic motion which repeats itself around a point known as equilibrium point and in this motion applied force is directly proportional to displacement but it acts opposite in direction. Also the superposition principle states that the two Simple harmonic motions can be added linearly.

Complete step by step answer:
For a SHM of both we would consider the general solution, for the first SHM amplitude is ${a_1}$ and let it vibrate with angular frequency $\omega $. Also, the initial phase of the first SHM is ${\alpha _1}$. So its displacement be given as,
${x_1} = {a_1}\sin (\omega t + {\alpha _1})$

Similarly for the second SHM, amplitude is ${a_2}$ and let it vibrate with angular frequency $\omega $.Also, let the initial phase of second SHM is ${\alpha _2}$.So its displacement be given as,
${x_2} = {a_2}\sin (\omega t + {\alpha _2})$
So according to superposition principle,
$x = {x_1} + {x_2}$
Putting values we get,
$x = {a_1}\sin (\omega t + {\alpha _1}) + {a_2}\sin (\omega t + {\alpha _2})$
$\Rightarrow x = {a_1}(\sin (\omega t)\cos {\alpha _1} + \cos (\omega t)\sin {\alpha _1}) + {a_2}(\sin (\omega t)\cos {\alpha _2} + \cos (\omega t)\sin {\alpha _2})$
$\Rightarrow x = \sin (\omega t)[{a_1}\cos {\alpha _1} + {a_2}\cos {\alpha _2}] + \cos (\omega t)[{a_1}\sin {\alpha _1} + {a_2}\sin {\alpha _2}]$
Let, ${a_1}\cos {\alpha _1} + {a_2}\cos {\alpha _2} = a\cos \alpha $
And, ${a_1}\sin {\alpha _1} + {a_2}\sin {\alpha _2} = a\sin \alpha $
So, $x = a\sin (\omega t)\cos \alpha + a\cos (\omega t)\sin \alpha $
$\therefore x = a\sin (\omega t + \alpha )$
This equation is analogous to Simple Harmonic motion with different amplitude but the period of motion is the same as the period is defined as, $T = \dfrac{{2\pi }}{\omega }$ and it is the same.

So the correct option is B.

Note: It should be remembered that Simple harmonic motion is different from periodic motion. So all harmonic motions are periodic but all periodic are not harmonic. For a periodic motion to be harmonic motion acceleration must be directly proportional to displacement and it must be opposite in direction. Real world examples are motion of pendulum etc.