Question

Three simple harmonic motions in the same direction, each of amplitude ${}^{\prime }{a}^{\prime }$ and periodic time $T$ are superposed. The first and second, and the second and third differ in phase from each other by $\pi /4$, with the first and third not being identical. Then

• A. the resultant motion is not simple harmonic
• B. the resultant amplitude is $(\sqrt{2}+1)a$
• C. the phase difference between the second $SHM$ and the resultant motion is zero
• D. the energy in the resultant motion is three times the energy in each separate $SHM$

Answer 1

Answer: (B), (C)

The correct options are
B the resultant amplitude is $(\sqrt{2}+1)a$
C the phase difference between the second $SHM$ and the resultant motion is zero

Let the simple harmonic motions be given by
$x_1=a\sin \left(2\pi \frac{t}{T}\right)$
$x_2=a\sin (2\pi \frac{t}{T}+\frac{\pi }{4})$
and
$x_3=a\sin (2\pi \frac{t}{T}+\frac{\pi }{2})$
Then the resultant periodic motion, by the principle of superposition is given by
$x=x_1+x_2+x_3$
$x=a\left(\sin \frac{2\pi t}{T}\right)+a\sin (\frac{2\pi t}{T}+\frac{\pi }{4})+a\sin (\frac{2\pi t}{T}+\frac{\pi }{2})$
$x=a[\sin \left(\frac{2\pi t}{T}\right)+a\sin (\frac{2\pi t}{T}+\frac{\pi }{2})]+a\sin (\frac{2\pi t}{T}+\frac{\pi }{4})$
$x=2a\sin (\frac{2\pi t}{T}+\frac{\pi }{4})\cos \frac{\pi }{4}+a\sin (\frac{2\pi t}{T}+\frac{\pi }{4})$
$=a(\sqrt{2}+1)\sin (\frac{2\pi t}{T}+\frac{\pi }{4})$
which is a simple harmonic motion with an amplitude $a(\sqrt{2}+1)$ and phase angle $\pi /4$ and the same period; it has the same period; it has the same phase as second $SHM$.
${\left[a\right(\sqrt{2}+1\left)\right]}^2(i.e.)\left[a^2\right(2+1+2\sqrt{2}\left)\right]=(3+2\sqrt{2})a^2$ which is greater than three times the energy of each separate $SHM$