Harmonics Question; prove that v max of a mass on a spring is given by 2 pi f A

[lost_in_phys]

The question is exactly this:

Prove that the maximum speed (Vmax) of a mass on a a spring is given by 2(pi)fA.

where f = frequency and A =Amplitude

We are given such formulas as:
f=[1 / 2(pi)]square_root (k /m) or
f=[1 / 2(pi)]square_root (a /-x)
the formula for total energy in a system (1/2*m*vsquared + 1/2*k*xsquared)

When I try to do it, I try to rearrange the formulas and get stuck in a giant maze of variables and such.

This question was asked before on this forum as part of another question but was never answered

 

[rootX]

Use energy equations!
What does k/m equals to?
Go back to the basics.

 

[lost_in_phys]

that doesn't really help me. Maybe mentioning which ones I should use would help, because I start to use some but they don't seem to get me where I want. It's like math class all over again with that darn unit with rearranging cos(x), sin(x), and tan(x), i hated that too.

 

[lost_in_phys]

nvm, i figured it out

 

[tascja]

can anyone elaborate on how he figured it out?

 

[rock.freak667]

can anyone elaborate on how he figured it out?

For SHM

$$v= \pm \omega \sqrt{A^2-x^2}$$



x is the displacement from the equilibrium position.At +A, you have max pe and 0 ke.
At -A you have max pe and 0 ke. At the equilibrium position you have 0pe and max ke (which means?)

 

[tascja]

im not really following because I've never seen that equation before; i understand the amplitude part where at equilibrium the mass has kinetic energy while before or after the equilibrium the mass will have potential energy.

 

[rock.freak667]

im not really following because I've never seen that equation before

The basic SHM eq'n: $a=- \omega^2 x$

$$a=\frac{dv}{dt}=v \frac{dv}{dx}$$


so that

$$v \frac{dv}{dx}= - \omega^2 x$$


and solve that, remembering that when x=A,v=0 and you'll arrive at the equation.

 

[tascja]

For SHM

$$v= \pm \omega \sqrt{A^2-x^2}$$



x is the displacement from the equilibrium position.At +A, you have max pe and 0 ke.
At -A you have max pe and 0 ke. At the equilibrium position you have 0pe and max ke (which means?)

oh okay so i read up on SHM and circular motion on the internet (not in my textbook) and it says that: Note that the in the SHM displacement equation is known as the angular frequency. It is related to the frequency (f) of the motion, and inversely related to the period (T):

$$\omega = 2 \pi f$$
so...
$$v= \pm \omega \sqrt{A^2-x^2}$$
$$v= 2 \pi f \sqrt{A^2-x^2}$$

and since velocity is at a maximum that means that displacement is 0 so x = 0 and then:

$$v= 2 \pi f \sqrt{A^2-0}$$
$$v= 2 \pi f A^2$$

right?
I just don't really understand what $$\omega$$ stands for?

 

[rock.freak667]

Note:$\sqrt{A^2}=A$


$\omega$ is the angular frequency i.e. the number of oscillations in $2 \pi$ seconds

 

[tascja]

oh yea i forgot to take that off, but thank you for all your help!

 

[Nocholas]

But if we want an answer without straying from the text (assuming you are using ILC's SPH4U funded by TVO), then we can go to Lesson 6, page 12.

Right at the top, it gives the equation T=2(pi)r/v
*If you know how to graph trig functions, or understand them even in the least, you know that the (r)adius is the same as the (A)mplitude

Thus: T= 2(pi)A/v
Isolating for v, we get: v=2(pi)A/T
But since T=1/f;
v=2(pi)Af

(v)oila