IIT JEE , AIEEE Forum - Physics , Mechanics

bright_mahendra

A body is executing SHM under the action of force whose maximum magnitude is 50N. Find the magnitude of force acting on the particle at the time when its energy is half kinetic and half potential.

nishant_jain

Given,

kA = 50N

1/2KX²=1/2mv²

hence , v=wx

which means ,

wAcos(wt)=wAsin(wt)

tan(wt)=1

wt=pie/4

Now,

x=A*sin(45)

A/1.414

we know,

kA=50

k=50/A

Magnitude of force = kx

(50/A) * (A/1.414)

=50/1.414

krishna.gopal

Right answer Nishant

bright_mahendra

Thanks nishant.

Asmita

Please tell me for what does each of the variables used in the sol. represent.

nishant_jain

Hi asmita,

Here are the meanings of each variable used...

k - constant (=m*w²)

A - Amplitude

x - variable for displacement of a particle

v - variable for velocity of a particle

w - anguar frequency

t - time

m - mass

Asmita

Thanks a lot.

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