MHT CET 2021 23rd September Evening Shift | Simple Harmonic Motion Question 72 | Physics

A bob of simple pendulum of mass 'm' perform $$\mathrm{SHM}$$ with amplitude '$$\mathrm{A}$$' and period 'T'. Kinetic energy of pendulum of displacement $$x=\frac{A}{2}$$ will be

$$\frac{2 \mathrm{~m} \pi^2 A}{3 \mathrm{~T}^2}$$

$$\mathrm{\frac{3 m \pi^2 A}{2 T}}$$

$$\frac{2 \mathrm{~m} \pi \mathrm{A}^2}{3 \mathrm{~T}}$$

$$\frac{2 \mathrm{~m} \pi^2 \mathrm{~A}^2}{2 \mathrm{~T}^2}$$

MHT CET 2021 23rd September Evening Shift

MCQ (Single Correct Answer)

-0

An object executes SHM along $$x$$-axis with amplitude $$0.06 \mathrm{~m}$$. At certain distance '$$\mathrm{x}$$' metre from mean position, it has kinetic energy $$10 \mathrm{~J}$$ and potential energy $$8 \mathrm{~J}$$. the distance '$$\mathrm{x}$$' will be

0.08 m

0.02 m

0.04 m

0.06 m

MHT CET 2021 23rd September Evening Shift

MCQ (Single Correct Answer)

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A body executes SHM under the action of force '$$\mathrm{F}_1$$' with time period '$$\mathrm{T}_1$$'. If the force is changed to '$$\mathrm{F_2}$$', it executes SHM with period '$$\mathrm{T_2}$$'. If both the forces '$$\mathrm{F_1}$$' and '$$\mathrm{F}_2$$' act simultaneously in the same direction on the body, its time period is

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

$$\frac{T_1 T_2}{\sqrt{T_1^2-T_2^2}}$$

$$\frac{\sqrt{T_1^2+\mathrm{T}_2^2}}{\mathrm{~T}_1 \mathrm{~T}_2}$$

$$\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}$$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

-0

A particle performing S.H.M. when displacement is '$$x$$', the potential energy and restoring force acting on it are denoted by '$$E$$' and '$$F$$' respectively. The relation between $$x, E$$ and $$F$$ is

$$\frac{2 E}{F}-x^2=0$$

$$\frac{2 \mathrm{E}}{\mathrm{F}}+\mathrm{x}^2=0$$

$$\frac{2 E}{F}+x=0$$

$$\frac{2 E}{F}-x=0$$

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