WB JEE 2008 | Simple Harmonic Motion Question 1 | Physics

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are K₁ and K₂ respectively. The period of vertical oscillations of the system will be

$${1 \over {8\pi }}\sqrt {{K_1} + {K_2}} $$

$$8\pi \sqrt {{{{K_1} + {K_2}} \over {{K_1}{K_2}}}} $$

$${\pi \over 2}\sqrt {{K_1} - {K_2}} $$

$${\pi \over 2}\sqrt {{{{K_1}} \over {{K_2}}}} $$

WB JEE 2025

MCQ (Single Correct Answer)

-0.25

A simple pendulum is taken at a place where its distance from the earth's surface is equal to the radius of the earth. Calculate the time period of small oscillations if the length of the string is 4.0 m . (Take $g=\pi^2 \mathrm{~ms}^{-2}$ at the surface of the earth.)

4 s

6 s

8 s

2 s

WB JEE 2025

MCQ (Single Correct Answer)

-0.25

The variation of displacement with time of a simple harmonic motion (SHM) for a particle of mass $m$ is represented by $y=2 \sin \left(\frac{\pi t}{2}+\phi\right) \mathrm{cm}$. The maximum acceleration of the particle is

$\frac{\pi}{2} \mathrm{~cm} / \mathrm{sec}^2$

$\frac{\pi}{2 m} \mathrm{~cm} / \mathrm{sec}^2$

$\frac{\pi^2}{2 m} \mathrm{~cm} / \mathrm{sec}^2$

$\frac{\pi^2}{2} \mathrm{~cm} / \mathrm{sec}^2$

WB JEE 2024

MCQ (Single Correct Answer)

-0.25

A particle of mass '$$m$$' moves in one dimension under the action of a conservative force whose potential energy has the form of $$U(x)=-\frac{\alpha x}{x^2+\beta^2}$$ where $$\alpha$$ and $$\beta$$ are dimensional parameters. The angular frequency of the oscillation is proportional to

$$\sqrt{\frac{\alpha^3}{\mathrm{~m} \beta^4}}$$

$$\sqrt{\frac{\alpha}{m \beta^4}}$$

$$\sqrt{\frac{\alpha}{\mathrm{m} \beta^3}}$$

$$\sqrt{\frac{\alpha}{m \beta^6}}$$

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