Q162 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics

Identify the equation that DOES NOT represent oscillatory motion.
($A$ is amplitude of motion)

  • A
    $y = A\sin\omega t + A\cos 2\omega t$
  • B
    $x = A\cos\left(\omega t - \dfrac{\pi}{4}\right)$
  • Correct answer
    C
    $y = A\sin\omega t + A\cos(\omega t + \pi/2)$
  • D
    $y = A\sin\left(\omega t - \dfrac{\pi}{4}\right)$

Solution

Let us analyze each option to determine if it represents oscillatory motion.

For the equation $y = A\sin\omega t + A\cos 2\omega t$, the motion is a superposition of two harmonic functions with commensurable frequencies. This represents a periodic and oscillatory motion.

For the equation $x = A\cos\left(\omega t - \dfrac{\pi}{4}\right)$, the motion is simple harmonic, which is inherently oscillatory.

For the equation $y = A\sin\omega t + A\cos(\omega t + \pi/2)$, we can simplify it using the trigonometric identity $\cos(\theta + \pi/2) = -\sin\theta$:
$y = A\sin\omega t - A\sin\omega t = 0$
Since the displacement $y$ is zero for all time $t$, the particle remains at rest at the origin. Therefore, this does not represent oscillatory motion.

For the equation $y = A\sin\left(\omega t - \dfrac{\pi}{4}\right)$, the motion is simple harmonic, which is oscillatory.

Thus, the equation that does not represent oscillatory motion is $y = A\sin\omega t + A\cos(\omega t + \pi/2)$.

Answer: $y = A\sin\omega t + A\cos(\omega t + \pi/2)$

Hence, the correct answer is option: 'C'.

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