Home
Class 12
PHYSICS
Statement-1 : A pendulum bob is susponde...

Statement-1 : A pendulum bob is susponded form a fixed support through a string of length `l`. It is executing simple harmonic motion of small amplitude. A person running with an acceleration a on the horizontal floor will oberve the time period of the pendulum to be `2pisqrt((l)/(sqrt(a^(2)+g^(2))))`.
Statement-2 : In frame attached to the person the bob will experience a psuedo force.

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explantion for Statement-1

B

Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explantion for Statement-1

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True.

Text Solution

AI Generated Solution

Promotional Banner

Similar Questions

Explore conceptually related problems

A particle executing simple harmonic motion with an amplitude 5 cm and a time period 0.2 s. the velocity and acceleration of the particle when the displacement is 5 cm is

The position velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2cm, 1ms^-1 and 10ms^-2 at a certain instant. Find the amplitude and the time period of the motion.

Consider the following statements. The total energy of a particles executing simple harmonic motion depends on its 1. amplitude 2. Period 3. displacement of these :

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally currect

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally correct

A body of mass m attached to one end of an ideal spring of force constant k is executing simple harmonic motion. Establish that the time - period of oscillation is T=2pisqrt(m//k) .

Statement-1 : The motion of a simple pendulum is not simple harmonic for large amplitudes. Statement-2 : The restoring torque on a simple pendulum about the point of suspension is proportional to sintheta , where theta is the angular displacement of the pendulum.

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration alpha , then the time period is given by T = 2pisqrt(((I)/(g))) where g is equal to

Masses M_(A) "and" M_(B) hanging from the ends of strings of lengths L_(A) "and" L_(B) are executing simple harmonic motions. If their frequencies are f_(A) = 2f_(B) , then

Calculate percentage error in determination of time period of a pendulum. T = 2pi (sqrt(l))/(g) where, l and g are measured with +-1% and +-2% .