MHT CET 2019 3rd May Morning Shift | Simple Harmonic Motion Question 123 | Physics

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be $T$. If the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$, the time period of the pendulum will be

$\frac{T}{\sqrt{3}}$

$\sqrt{3} \frac{T}{2}$

$\sqrt{3} T$

$\frac{T}{3}$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

In damped SHM, the SI unit of damping constant is

$\frac{\mathrm{N}}{\mathrm{s}}$

$\frac{\mathrm{kg}}{\mathrm{s}}$

$\frac{\mathrm{kg}}{\mathrm{m}}$

$\frac{N}{m}$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

The total energy of a simple harmonic oscillaior is proportional to

square of this ampitude

square root of displacement

amplitude

frequency

MHT CET 2019 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

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

$\frac{\sqrt{3}}{2 \pi} s$

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

$\frac{1}{2 \pi \sqrt{3}} \mathrm{~s}$

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