TG EAPCET 2024 (Online) 9th May Evening Shift | Simple Harmonic Motion Question 4 | Physics

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

$800 \times 10^{-6} \mathrm{Jm}^{-2}$

$1600 \times 10^{-6} \mathrm{Jm}^{-2}$

$3200 \times 10^{-6} \mathrm{Jm}^{-2}$

$6400 \times 1^{-6} \mathrm{Jm}^{-2}$

TG EAPCET 2024 (Online) 9th May Morning Shift

MCQ (Single Correct Answer)

-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

$40 \%$

$20 \%$

$75 \%$

$50 \%$

TS EAMCET 2023 (Online) 12th May Evening Shift

MCQ (Single Correct Answer)

-0

A clock is designed based on the oscillation of a spring-block system suspended vertically in the absence of air-resistance. Assume it shows the correct time when a spring of stiffness $k$ and block is mass $m$ are used. If the block is replaced by another block of mass $4 m$, choose the correct option

The clock runs slow by 0.5 s for every second

The clock runs fast by 0.5 s for every one second

The clock runs fast by 1 s for every one second

The clock runs slow by 1 s for every one second

TS EAMCET 2023 (Online) 12th May Morning Shift

MCQ (Single Correct Answer)

-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

$2 \sqrt{6} \mathrm{~cm}$

$\frac{2}{\sqrt{6}} \mathrm{~cm}$

$\sqrt{2} \mathrm{~cm}$

$\frac{6}{\sqrt{2}} \mathrm{~cm}$

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