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

In summary, the maximum speed (Vmax) of a mass on a spring can be found using the formula 2(pi)fA, where f is the frequency and A is the amplitude. This can be derived from the equations for frequency and total energy of a system. The angular frequency, denoted as \omega, is the number of oscillations in 2(pi) seconds and can also be expressed as 2(pi)f. Using the equation T=2(pi)A/v, we can isolate for v and arrive at the formula v=2(pi)Af, where A is the amplitude and f is the frequency.
  • #1
lost_in_phys
8
0
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
 
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  • #2
Use energy equations!
What does k/m equals to?
Go back to the basics.
 
  • #3
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.
 
  • #4
nvm, i figured it out
 
  • #5
can anyone elaborate on how he figured it out?
 
  • #6
tascja said:
can anyone elaborate on how he figured it out?

For SHM

[tex]v= \pm \omega \sqrt{A^2-x^2}[/tex]



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?)
 
  • #7
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.
 
  • #8
tascja said:
im not really following because I've never seen that equation before

The basic SHM eq'n: [itex]a=- \omega^2 x[/itex]

[tex]a=\frac{dv}{dt}=v \frac{dv}{dx}[/tex]


so that

[tex]v \frac{dv}{dx}= - \omega^2 x[/tex]


and solve that, remembering that when x=A,v=0 and you'll arrive at the equation.
 
  • #9
rock.freak667 said:
For SHM

[tex]v= \pm \omega \sqrt{A^2-x^2}[/tex]



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):

[tex] \omega = 2 \pi f[/tex]
so...
[tex]v= \pm \omega \sqrt{A^2-x^2}[/tex]
[tex]v= 2 \pi f \sqrt{A^2-x^2}[/tex]

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

[tex] v= 2 \pi f \sqrt{A^2-0}[/tex]
[tex] v= 2 \pi f A^2 [/tex]

right? :confused:
I just don't really understand what [tex] \omega [/tex] stands for?
 
  • #10
Note:[itex]\sqrt{A^2}=A[/itex]


[itex]\omega[/itex] is the angular frequency i.e. the number of oscillations in [itex]2 \pi[/itex] seconds
 
  • #11
oh yea i forgot to take that off, but thank you for all your help!
 
  • #12
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
 

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

1. What is the equation for finding the maximum velocity of a mass on a spring?

The maximum velocity (vmax) of a mass on a spring is given by the formula vmax = 2πfA, where f is the frequency of the oscillation and A is the amplitude of the oscillation. This equation is known as the harmonic motion equation and is derived from the principles of simple harmonic motion.

2. How is the equation for maximum velocity derived?

The equation for maximum velocity is derived from the principles of simple harmonic motion, which states that the restoring force of a spring is directly proportional to the displacement of the mass from its equilibrium position. By applying Newton's second law of motion and solving for velocity, we arrive at the equation vmax = 2πfA.

3. What is the significance of the frequency and amplitude in the equation for maximum velocity?

The frequency (f) represents the number of oscillations per unit time and is measured in Hertz (Hz). The amplitude (A) is the maximum displacement of the mass from its equilibrium position. These two variables determine the speed at which the mass moves and the distance it covers during each oscillation.

4. How does the maximum velocity change with a change in frequency or amplitude?

The maximum velocity is directly proportional to both the frequency and amplitude. This means that as the frequency or amplitude increases, the maximum velocity also increases. Conversely, a decrease in frequency or amplitude results in a decrease in maximum velocity.

5. Can the equation for maximum velocity be used for all types of harmonic motion?

The equation vmax = 2πfA can be used for any type of harmonic motion, as long as the motion is periodic and follows the principles of simple harmonic motion. This includes systems such as a mass on a spring, a pendulum, and a vibrating string.

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