Posted by **Pronab** on Thursday, November 18, 2010 at 10:39am.

A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t=0 it is at position x=5 cm going towards positive x direction . Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4s.

- physics -
**bobpursley**, Thursday, November 18, 2010 at 1:22pm
A=10cm

w=2PIf=2PI/T= PI/3

f(t)=Acos(wt+Phi)

f(0)=10cm*cos(0+phi)=5 >>>phi=PI/3 radians

f(t)=10cm*cos(wt+PI/3)

f'=-10cm*w*sin(wt+PI/3)=velocity

f"=-10cm*w^2*cos(PI*t/3+PI/3)=acceleration

- physics -
**Pronab**, Saturday, November 20, 2010 at 7:47am
Q 2) A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?

- physics -
**surbhi**, Sunday, December 6, 2015 at 11:20pm
X=(10cm)sin(2¦Ð/6st+¦Ð/6),~11cm s

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