MHT CET 2024 2nd May Evening Shift | Simple Harmonic Motion Question 34 | Physics

A spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $\mathrm{T}_2$. The ratio of $T_2 / T_1$ is

$1: 2$

$1: \sqrt{2}$

$\sqrt{2}: 1$

$2: 1$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

For a body performing simple harmonic motion, its potential energy is $\mathrm{E}_{\mathrm{x}}$ at displacement x and $\mathrm{E}_{\mathrm{y}}$ at displacement y from mean position. The potential energy $E_0$ at displacement $(x+y)$ is

$\sqrt{\mathrm{E}_{\mathrm{x}}^2+\mathrm{E}_{\mathrm{y}}^2}$

$\sqrt{E_x-E_y}$

$E_x+E_y$

$E_x+E_y+2 \sqrt{E_x E_y}$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

The displacement of a particle performing S.H.M. is given by $Y=A \cos [\pi(t+\phi)]$. If at $\mathrm{t}=0$, the displacement is $\mathrm{y}=2 \mathrm{~cm}$ and velocity is $2 \pi \mathrm{~cm} / \mathrm{s}$, the value of amplitude $A$ in cm is

$\sqrt{2}$

$2 \sqrt{2}$

$\frac{1}{\sqrt{2}}$

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

A particle is performing simple harmonic motion and if the oscillations are Camped oscillations then the angular frequency is given by

$\sqrt{\frac{k}{m}+\left(\frac{b}{2 m}\right)^2}$

$\frac{\mathrm{k}}{\mathrm{m}}+\left(\frac{\mathrm{b}}{2 \mathrm{~m}}\right)^2$

$\sqrt{\frac{k}{m}-\left(\frac{b}{2 m}\right)^2}$

$\frac{\mathrm{k}}{\mathrm{m}}-\left(\frac{\mathrm{b}}{2 \mathrm{~m}}\right)^2$

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