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CBSE Class 11-science Answered

A particle oscillating harmonically with an amplitue of 1.5 cm has maximum energy of 0.25 x 10-6 J. At what displacement from the equilibrium position will the particle be acted upon by a force of 2.5 x 10-5 N?
Asked by Yash Mishra | 11 Feb, 2014, 03:38: AM
answered-by-expert Expert Answer
Answered by | 11 Feb, 2014, 12:48: PM

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CBSE 11-science - Physics
.A SHM is expressed by the equation x = A cos (ωt + φ) and the phase angle φ =0. Draw graphs to show variation of displacement, velocity and acceleration for one complete curve in SHM.
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CBSE 11-science - Physics
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the time period of simple pendulum is 2 second it can go from equilium position at the maximum distance of 5 cm if at the start of the motion the pendulum in the position of maximum displacement towards the right of equilium position then write the different equation.
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CBSE 11-science - Physics
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CBSE 11-science - Physics
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CBSE 11-science - Physics
In Shm where velocity is maximum or zero    
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CBSE 11-science - Physics
How does the change in the center of mass affect the time period of Bob? explain with diagram
Asked by polyrelation | 04 Dec, 2019, 10:35: AM
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CBSE 11-science - Physics
A small body of mass 0.10kg is undergoing simple harmonic motion of simplitude 1.0 meter and period of 0.20 sec the maximum value of the force acting on it
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CBSE 11-science - Physics
  A uniform circular disc of mass 12 kg is held by 2 identical springs when disc is pressed down slightly and released it executes shm with time period of2 s the force constant of each spring is [2 spring in parallel]n/
Asked by SELVA | 06 Jul, 2019, 03:25: PM
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