MHT CET 2024 4th May Morning Shift | Simple Harmonic Motion Question 27 | Physics

A simple pendulum of length $l_1$ has time period $\mathrm{T}_1$. Another simple pendulum of length $l_2\left(l_1>l_2\right)$ has time period $T_2$. Then the time period of the pendulum of length $\left(l_1-l_2\right)$ will be

$T_1-T_2$

$\sqrt{\frac{T_1}{T_2}}$

$\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}$

$\sqrt{\frac{T_2}{T_1}}$

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

-0

Two bodies A and B of equal mass are suspended from two separate massless springs of spring constants $\mathrm{K}_1$ and $\mathrm{K}_2$ respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of $B$ to that of $A$ is

$\frac{\mathrm{K}_1}{\mathrm{~K}_2}$

$\frac{\mathrm{K}_2}{\mathrm{~K}_1}$

$\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}$

$\sqrt{\frac{\mathrm{K}_2}{\mathrm{~K}_1}}$

MHT CET 2024 3rd May Evening Shift

MCQ (Single Correct Answer)

-0

The period of a simple pendulum gets doubled when

its length is doubled.

its length is made four times.

its length is made half.

the mass of the bob is doubled.

MHT CET 2024 3rd May Evening Shift

MCQ (Single Correct Answer)

-0

Frequency of a particle performing S.H.M. is 10 Hz . The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

$\frac{1}{\pi} \mathrm{~m} / \mathrm{s}$

$\frac{1}{2 \pi} \mathrm{~m} / \mathrm{s}$

$\frac{1}{4 \pi} \mathrm{~m} / \mathrm{s}$

$2 \pi \mathrm{~m} / \mathrm{s}$

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