A particle executes I inear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is In

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Q: 60 (NDA-II/2016)
A particle executes I inear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is In

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,5,5,10,0,0,0

keywords: 

{'simple harmonic motion': [0, 0, 4, 4], 'acceleration': [0, 0, 2, 8], 'amplitude': [1, 0, 7, 8], 'velocity': [0, 2, 2, 6], 'particle': [0, 2, 8, 30], 'particle executes': [0, 0, 0, 1], 'seconds': [3, 3, 8, 6], 'mean position': [0, 0, 3, 1], 'magnitudes': [0, 0, 0, 2], 'time period': [1, 0, 1, 12], 'cm': [2, 0, 7, 20]}

This question is about a particle executing linear simple harmonic motion with an amplitude of 2 cm. The particle is at a distance of 1 cm from the mean position, and at this point, the magnitudes of the velocity and the acceleration are equal. We need to find the time period of the motion.

In simple harmonic motion, the equation for displacement as a function of time is given by x = A*cos(ωt), where A is the amplitude and ω is the angular frequency.

To find the time period, we need to determine the value of ω. The velocity of the particle at any point in simple harmonic motion is given by v = -A*ω*sin(ωt), and the acceleration is given by a = -A*ω^2*cos(ωt).

From the given information, we know that the magnitudes of velocity and acceleration are equal when the particle is at a distance of 1 cm from the mean position. This means that |v| = |a|, which gives us:

|-A*ω*sin(ωt)| = |-A*ω^2*cos(ωt)|

Taking the ratio of the two equations, we have:

|sin(ωt)|/|cos(ωt)| =