List of top Physics Questions on Simple Harmonic Motion asked in AP EAPCET

Two blocks each of mass \(m\) are connected to a spring of spring constant \(K\). If both are given velocity \(V\) in opposite directions as shown in the figure, then the maximum elongation of the spring is center

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A particle is executing simple harmonic motion with time period \(T\) and amplitude \(A\). The distance travelled by the particle in \(\frac{T}{12}\) time starting from rest is:

Which of the following statements are correct for a particle in simple harmonic motion? I. Its velocity-displacement graph is a parabola. II. Its velocity-time graph is sinusoidal. III. Its velocity-acceleration graph is an ellipse.

The potential energy of a particle of mass 1 kg which is in simple harmonic motion along X-axis is given by $U(x)=4(1-\cos 2x)$. The time period of oscillations is:

The energy of a particle executing Simple Harmonic Motion (SHM) is given by E = Ax^2 + BV^2. Here 'x' is the displacement of the particle from its mean position, 'V' is its velocity at 'x', and A and B are positive constants. The maximum velocity of the particle is:

The structural characteristics of Simple Harmonic Motion (SHM) require that the acceleration of a particle is directly proportional to its displacement from the mean position and is directed:

The energy of a particle executing SHM is given by E = Ax^2 + BV^2. The INCORRECT statement is:

The springs are connected to the blocks as shown in figures A and B. When the blocks are slightly displaced and released, they oscillate with time periods T_A and T_B respectively. Then, the value of T_AT_B is:

A particle executes two simple harmonic motions along mutually perpendicular axes, given by \( x = A \sin(\omega_1 t) \) and \( y = B \cos(\omega_2 t) \) where \( A \neq B \) and \( \omega_1 = \omega_2 \). Which of the following best describes the resultant motion of the particle?

The maximum kinetic energy of a pendulum executing simple harmonic motion is E. If the length of the pendulum is doubled and the amplitude of motion is halved, then the maximum kinetic energy of the pendulum is:

The equation of motion of a particle executing simple harmonic motion is given by \( X=\sqrt{2} [1.2 \sin 2t - 1.6 \cos 2t] \), where \( X \) is displacement in metre and \( t \) is time in second. The velocity of the particle at a time of 0.125 s is:

If the minimum time taken by a particle executing simple harmonic motion to move from extreme position to a point at a displacement of 86.6% of the amplitude is T, then the minimum time taken by the particle to move from mean position to a point at a displacement of 86.6% of the amplitude is:

For a body in simple harmonic motion. Potential energy (PE), Kinetic energy (KE), Total energy (TE) are measured as a function of displacement x. Which of the following statements is true?

A particle executes simple harmonic motion with time period 'T' and amplitude 'a'. The magnitude of average velocity of the particle over the time interval during which it travels a distance from extreme position to a/2 is:

A particle starting from the origin executes simple harmonic motion along x-axis. Its velocity at any instant t is given by $v_{x}=22 \cos(\frac{\pi t}{2}) \text{ cm s}^{-1}$. The total distance covered by the particle in time $t=4.5 \text{ sec}$ is

In simple harmonic motion, at mean position

A particle executes SHM with amplitude \( 0.1 \text{ m} \) and angular frequency \( 10 \text{ rad/s} \). Maximum acceleration is:

Two SHMs are given by: \[ x_1 = A \sin \omega t, \] \[ x_2 = A \cos \omega t. \] Phase difference is:

A particle in SHM has amplitude \( 2 \text{ m} \) and total energy \( 8 \text{ J} \). Its potential energy at displacement \( 1 \text{ m} \) is:

A mass of \( 1 \text{ kg} \) is attached to a spring of force constant \( 100 \text{ N/m} \). Time period is:

A mass of 1 kg is attached to a spring of force constant 100 N/m. Time period is:

A particle is executing SHM of amplitude \(4\text{ cm}\) and time period \(4\text{ s}\). The time taken by it to move from positive extreme position to half the amplitude is

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $ E $ is the total energy):

Assertion (A): In S.H.M., kinetic and potential energies become equal when the distance is \[ \frac{1}{\sqrt{2}} \] times the amplitude. Reason (R): The potential energy of a particle executing S.H.M. is periodic with time and is maximum at the extreme displacement.

Position of a \(3\ \text{kg}\) mass moving along the \(x\)-axis is given by \(x=0.3\cos(\omega t)\ \text{m}\). If \(K(t)\) denotes the kinetic energy at time \(t\), then the value of \(\dfrac{K\left(\frac{\pi}{6\omega}\right)}{K\left(\frac{\pi}{3\omega}\right)}\) is