Simple Harmonic Motion (1 Viewer)

[rawker]

Can someone help please

A particle is moving in SHM with displacement, in metres, over time t seconds, given by x = 2 sin 6 t.

Find the equation of the velocity of the paticle in terms of t.

...........
x= 2sin6t
x'=12cos6t
x"=-72sin6t=-36x
...........

 

[insert-username]

Can someone help please

A particle is moving in SHM with displacement, in metres, over time t seconds, given by x = 2 sin 6 t.

Find the equation of the velocity of the paticle in terms of t.

...........
x= 2sin6t
x'=12cos6t
x"=-72sin6t=-36x
...........

I'm not sure why you need help, because you've got the answer right there. Velocity is dx/dt, and dx/dt is the derivative of 2sin6t is 12cos6t.


I_F

 

[rawker]

Eeek. I meant in terms of x not t sorry.

 

[insert-username]

Okey. For any particle moving in Simple Harmonic Motion (you must learn this formula):

(x')² = n²(a² - x²), where a = amplitude and n = coefficient of t

So:

(x')² = 36(4 - x²)

x' = 6√(4 - x²)

That should be right now.


I_F

 

[Mountain.Dew]

x"=-72sin6t=-36x

its also worth nothing to use the derived equation:

x'' = d^2(x)/dt^2 = d(1/2v^2)/dx

then, u have d(1/2v^2)/dx = -36x ==> from there, integrate, and realise that when v = 0, x = a, a = amplitude. ==> then u get ur value for 'c', ur constant when u got ur primitive function from d(1/2v^2)/dx, and so u will anticipate the same result that insert-username has gladly provided us with:
x' = 6√(4 - x²), where x' = dx/dt