1
TG EAPCET 2025 (Online) 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The force ( $F$ in newton) acting on a particle of mass 90 g executing simple harmonic motion is given by $F+0.04 \pi^2 y=0$, where $y$ is displacement of the particle in metre. If the amplitude of the particle is $\frac{6}{\pi} \mathrm{~m}$, then the maximum velocity of the particle is

A

$6 \mathrm{~ms}^{-1}$

B

$2 \mathrm{~ms}^{-1}$

C

$8 \mathrm{~ms}^{-1}$

D

$4 \mathrm{~ms}^{-1}$

2
TG EAPCET 2025 (Online) 4th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the amplitudes of a damped harmonic oscillator at times $t=0, t_1$ and $t_2$ are $A_0, A_1$ and $A_2$ respectively, then the amplitude of the oscillator at a time of $\left(t_1+t_2\right)$ is

A

$\frac{A_0+A_1+A_2}{3}$

B

$\frac{A_2 A_0}{A_1}$

C

$\frac{A_1 A_0}{A_2}$

D

$\frac{A_1 A_2}{A_0}$

3
TG EAPCET 2025 (Online) 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

At a given place, to increase the number of oscillations made by a simple pendulum in one minute from 72 to 90 , the length of the pendulum is to be decreased by

A

$64 \%$

B

$36 \%$

C

$50 \%$

D

$56 \%$

4
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the amplitude of a damped harmonic oscillator becomes half of its initial amplitude in a time of 10 s , then the time taken for the mechanical energy of the oscillator to become half of its initial mechanical energy is

A

2.5 s

B

20 s

C

10 s

D

5 s

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