Rankers Physics

miscellaneous: Practice Problem & Solution

The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along: (1990)
Circle
Figure of eight
Straight line
Ellipse

Solution Explained:

To solve this problem, we apply the core principles of miscellaneous. Understanding the underlying formula is key to arriving at the correct answer below:

Let $x = A\sin(\omega t)$ and $y = B\sin(\omega t + \pi) = -B\sin(\omega t)$.\nThen $y/x = -B/A \implies y = -(B/A)x$.\nThis represents the equation of a straight line.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): In simple harmonic motion total mechanical energy can be negative also. Reason (R): Potential energy is always negative and if it is greater than kinetic energy total mechanical energy will be negative.  
2. Assertion (A): For a particle performing SHM, its speed decreases as it goes away from the mean position. Reason (R): In SHM, the acceleration is always opposite to the velocity of the particle.  
3. The total energy of particle performing S.H.M. depend on: (2001)
4. Assertion (A): Displacement-time equation of a particle moving along \(x\)-axis is \(x = 4 + 6 sin\omega t\). Under this situation, motion of particle is not simple harmonic. Reason (R): \(\frac{d^2x}{dt^2}\) for the given equation is not proportional to \(-x\).  
5. A particle is executing S.H.M. along a straight line. Its velocities at distances $ x_1 $ and $ x_2 $ from the mean position are $ v_1 $ and $ v_2 $, respectively. Its time period is: (2015)