Understanding the Time-Varying Induced EMF in a Spring-Magnetic Field System

[XuryaX]

A Rectangular wire frame, is placed in a uniform magnetic field directed upward and normal to the screen. A side AB is connected to the spring . The spring is stretched and released when the part has come to a point A'B'(t=0). How would the induced emf vary with time? - Graph & Equation

I tried using E=Blv.
deriving v from
f=kx
m * a = kx
v/t=kx/m
v=kxt/m

But it is probably not right. I don't have the solution of the book with the question. SO i am kinda confused.

Thanks
-XuryaX

 

[Abhishekdas]

m * a = kx
v/t=kx/m
v=kxt/m

Hold on...a=dv/dt NOT v/t ... your equation is correct only for uniformly accelerated motion with zero initial velocity but this case is not uniformly accelerated motion...So try getting an expression for simple harmonic motion of the rod in a trigonometric form and try to get the velocity of the rod from there at any time t...

 

[XuryaX]

Still it seems the same.

dv/dt=kx/m

dv=(kx/m)dt

Integrating

v= kxt/m

If we take dv/dt as d^2x/dt^2
It will be different though.

If we take that.
Then integrating to find x and then differentiating it to get v gives.

v=2kt/3x(squared)m

 

[Abhishekdas]

No...i said you have to get a trigonometric equation for the simple harmonic motion(since its a spring)...Like in this case it can be x=Acoswt...If w is given the sum can be solved...

 

[Abhishekdas]

dv/dt=kx/m

dv=(kx/m)dt

Integrating

v= kxt/m

This is wrong because you are treating x as a constant...Velocity is a function of x so the integration is not so simple...

F=-kx

so m*d²x/dt² = -kx ...this represents a simple harmonic motion so you need to proceed using trigonometric equations...