MHT CET 2023 14th May Morning Shift | Simple Harmonic Motion Question 4 | Physics

A simple pendulum of length '$$l$$' and a bob of mass '$$\mathrm{m}$$' is executing S.H.M. of small amplitude '$$A$$'. The maximum tension in the string will be ($$\mathrm{g}=$$ acceleration due to gravity)

$$2 \mathrm{~mg}$$

$$\mathrm{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)^2\right]$$

$$\mathrm{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]^2$$

$$m g\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

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The displacement of a particle executing S.H.M. is $$x=\mathrm{a} \sin (\omega t-\phi)$$. Velocity of the particle at time $$\mathrm{t}=\frac{\phi}{\omega}$$ is $$\left(\cos 0^{\circ}=1\right)$$

$$\omega \cos \phi$$

$$\mathrm{a} \omega$$

$$\omega \cos 2 \phi$$

$$-\mathrm{a} \omega \cos 2 \phi$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

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The bob of simple pendulum of length '$$L$$' is released from a position of small angular displacement $$\theta$$. Its linear displacement at time '$$\mathrm{t}$$' is ( $$\mathrm{g}=$$ acceleration due to gravity)

$$L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \sin \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \cos \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \sin \left[\sqrt{\frac{g}{L}} \cdot t\right]$$

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

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Under the influence of force $$F_1$$ the body oscillates with a period $$T_1$$ and due to another force $$F_2$$ body oscillates with period $$T_2$$. If both forces acts simultaneously, then the resultant period is (consider displacement is same in all three cases)

$$T=\sqrt{\frac{T_1^2+T_2^2}{T_1^2 T_2^2}}$$

$$T=\sqrt{\frac{T_1^2 T_2^2}{T_1^2+T_2^2}}$$

$$T=\sqrt{\frac{T_1^2}{T_2^2}}$$

$$T=\sqrt{T_1^2+T_2^2}$$

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