Embibe Experts Solutions for Exercise 2: EXERCISE 12.2

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Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Exercise 2: EXERCISE 12.2

Attempt the free practice questions from Exercise 2: EXERCISE 12.2 with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Physics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Exercise 2: EXERCISE 12.2 with Hints & Solutions

MEDIUM
JEE Main/Advance
IMPORTANT

When the maximum KE of a simple pendulum is K, then what is its displacement (in terms of amplitude a), when its KE is K2 ?

HARD
JEE Main/Advance
IMPORTANT

The kinetic energy of a simple harmonic motion when the particle is half way to its end point is (where E0 is the total energy)

EASY
JEE Main/Advance
IMPORTANT

For a particle executing simple harmonic motion, the kinetic energy K is given by, K=K0cos2ωt. The maximum value of potential energy is

EASY
JEE Main/Advance
IMPORTANT

Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy?

EASY
JEE Main/Advance
IMPORTANT

If a simple pendulum of length L has maximum angular displacement α, then the maximum kinetic energy of bob of mass M is

EASY
JEE Main/Advance
IMPORTANT

If U is the potential energy of an oscillating particle and F is the force acting on it at a given instant, which of the following is true? (Take x be the displacement of particle from mean position)

MEDIUM
JEE Main/Advance
IMPORTANT

A particle starts oscillating in simple harmonic motion from its equilibrium position with time period T. The ratio of KE and PE of the particle at time t=T12 is

MEDIUM
JEE Main/Advance
IMPORTANT

Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy?