MHT CET 2021 23rd September Evening Shift | Simple Harmonic Motion Question 74 | Physics

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

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

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

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

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

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

-0

A body is performing S.H.M. of amplitude 'A'. The displacement of the body from a point where kinetic energy is maximum to a point where potential energy is maximum, is

zero

$$\pm \mathrm{A}$$

$$\pm \frac{\mathrm{A}}{2}$$

$$\pm \frac{\mathrm{A}}{4}$$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

-0

A particle excuting S.H.M starts from the mean position. Its amplitude is 'A' and time period '$$\mathrm{T}$$' At what displacement its speed is one-fourth of the maximum speed?

$$\frac{\mathrm{A}}{\sqrt{15}}$$

$$\frac{\mathrm{A}}{4}$$

$$\frac{4 \mathrm{~A}}{15}$$

$$\frac{\mathrm{A} \sqrt{15}}{40}$$

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