AP EAPCET 2022 - 5th July Morning Shift | Simple Harmonic Motion Question 21 | Physics

A particle is executing simple harmonic motion with an instantaneous displacement $$x=A \sin ^2\left(\omega t-\frac{\pi}{4}\right)$$. The time period of oscillation of the particle is

$$\frac{2 \pi}{\omega}$$

$$\frac{\pi}{\omega}$$

$$\frac{\pi}{2 \omega}$$

$$\frac{\omega}{2 \pi}$$

AP EAPCET 2022 - 5th July Morning Shift

MCQ (Single Correct Answer)

-0

If the amplitude of a lightly damped oscillator decreases by $$1.5 \%$$ then the mechanical energy of the oscillator lost in each cycle is

1.5%

0.75%

AP EAPCET 2022 - 4th July Evening Shift

MCQ (Single Correct Answer)

-0

A body is executing S.H.M. At a displacement $$x$$ its potential energy is 9 J and at a displacement $$y$$ its potential energy is 16 J . The potential energy at displacement $$(x+y)$$ is

25 J

5 J

49 J

7 J

AP EAPCET 2022 - 4th July Morning Shift

MCQ (Single Correct Answer)

-0

A hydrometer executes simple harmonic motion when it is pushed down vertically in a liquid of density $$\rho$$. If the mass of hydrometer is $$m$$ and the radius of the hydrometer tube is $$r$$, then the time period of oscillation is

$$T=2 \pi \sqrt{\frac{m}{\pi^2 \rho g}}$$

$$T=2 \pi \sqrt{\frac{\pi \pi^2 \rho g}{m}}$$

$$T=\frac{1}{2 \pi} \sqrt{\frac{m}{\pi \pi^2 \rho g}}$$

$$T=\frac{1}{2 \pi} \sqrt{\frac{\pi \pi^2 \rho g}{m}}$$

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