1
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
A particle of mass 4 mg is executing simple harmonic motion along $X$-axis with an angular frequency of $40 \mathrm{rad} \mathrm{s}^{-1}$. If the potential energy of the particle is $V(x)=a+b x^2$, where $V(x)$ is in joule and $x$ is in metre, then the value of $b$ is
A
$800 \times 10^{-6} \mathrm{Jm}^{-2}$
B
$1600 \times 10^{-6} \mathrm{Jm}^{-2}$
C
$3200 \times 10^{-6} \mathrm{Jm}^{-2}$
D
$6400 \times 1^{-6} \mathrm{Jm}^{-2}$
2
TG EAPCET 2024 (Online) 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
In a time $t$ amplitude of vibrations of a damped oscillator becomes half of its initial value, then the mechanical energy of the oscillator decreases by
A
$40 \%$
B
$20 \%$
C
$75 \%$
D
$50 \%$
3
TS EAMCET 2023 (Online) 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of 4 cm from the mean position is $1 / 3$ rd of the maximum kinetic energy. The amplitude of the motion is
A
$2 \sqrt{6} \mathrm{~cm}$
B
$\frac{2}{\sqrt{6}} \mathrm{~cm}$
C
$\sqrt{2} \mathrm{~cm}$
D
$\frac{6}{\sqrt{2}} \mathrm{~cm}$
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