AP EAPCET 2025 - 23rd May Morning Shift | Simple Harmonic Motion Question 9 | Physics

A body of mass 4 kg attached to a spring of force constant $64 \mathrm{Nm}^{-1}$ executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is

$\frac{\pi}{3} \mathrm{~s}$

$\frac{\pi}{2} \mathrm{~s}$

$\pi \mathrm{s}$

$\frac{3 \pi}{2} \mathrm{~s}$

AP EAPCET 2025 - 23rd May Morning Shift

MCQ (Single Correct Answer)

-0

A particle is executing simple harmonic motion with amplitude $A$. At a distance ' $x$ ' from the mean position, when the particle is moving towards extreme position it receives a blow in the direction of motion which instantaneously doubles its velocity. The new amplitude of the particle is

(Frequency is constant during the motion)

$A$

$\sqrt{A^2-X^2}$

$\sqrt{2 A^2-3 x^2}$

$\sqrt{4 A^2-3 x^2}$

AP EAPCET 2025 - 22nd May Evening Shift

MCQ (Single Correct Answer)

-0

If the displacement ' $x$ ' of a body in motion in terms of time ' $t$ ' is given by $x=A \sin (\omega t+\theta)$, then the minimum time at which the displacement becomes maximum is

$\left[\frac{\pi}{2 \omega}-\frac{\theta}{\omega}\right]$

$\left[\frac{2 \omega}{\pi}-\frac{\omega}{\theta}\right]$

$\left[\frac{\pi}{\omega}-\frac{1}{\omega}\right]$

$\left[\frac{\omega}{\pi}-\frac{\omega}{\pi^2}\right]$

AP EAPCET 2025 - 22nd May Evening Shift

MCQ (Single Correct Answer)

-0

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively $5 \mathrm{~ms}^{-1}$ and $10 \mathrm{~ms}^{-2}$, then the time period of oscillation of the particle is

$\pi \mathrm{s}$

$2 \pi \mathrm{~s}$

2 s

1 s

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