EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 16 - Simple Harmonic Motion>AMU-AT (B.Tech.) 2017>Q 2

EASY

AMU-AT (B.Tech.)

IMPORTANT

Earn 100

A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = k t^{2}, (k = 1 m s^{- 2})$ where $y$ is the vertical displacement. The time
period now becomes $T_{2}$ .

The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}} (g = 10 m s^{- 2})$ is

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Important Questions on Simple Harmonic Motion

EASY

AMU-AT (B.Tech.)

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 16 - Simple Harmonic Motion>AMU-AT (B.Tech.) 2018>Q 1

The time-period of a physical pendulum is

2 π \sqrt{\frac{I}{m g d}},

where

I

is the moment of inertia of the pendulum about the axis of rotation and

d

is perpendicular distance between the axis of rotation and the centre of mass of the pendulum. A circular ring hangs from a nail on a wall. The mass of the ring is

3 kg

and its radius is

20 cm

. If the ring is slightly displaced, the time of resulting oscillations will be

EASY

AMU-AT (B.Tech.)

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 16 - Simple Harmonic Motion>AMU-AT (B.Tech.) 2017>Q 2

A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = k t^{2}, (k = 1 m s^{- 2})$ where $y$ is the vertical displacement. The time
period now becomes $T_{2}$ .

The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}} (g = 10 m s^{- 2})$ is