How Can Impulse-Momentum Theorem Be Applied to Springs?

[Thomas Brown]

1. A mass M is connected to a wall by an ideal spring. The mass is on a frictionless surface. The mass is pushed toward the wall, compressing the spring by a distance X. Use the impulse-momentum theorem to demonstrate that the mass will reach a maximum velocity of X * (K / M)^1/2.

2. J = delta P, P = MV

3. J = delta P
Fn * t = MV - 0
-kx * t = MV What is t supposed to be? How can I finish this problem?

 

[Student100]

1. A mass M is connected to a wall by an ideal spring. The mass is on a frictionless surface. The mass is pushed toward the wall, compressing the spring by a distance X. Use the impulse-momentum theorem to demonstrate that the mass will reach a maximum velocity of X * (K / M)^1/2.

2. J = delta P, P = MV

3. J = delta P
Fn * t = MV - 0
-kx * t = MV What is t supposed to be? How can I finish this problem?

What is $\frac{\Delta v}{\Delta t}$ and what would the question wanting the maximum imply?

 

[Thomas Brown]

What is $\frac{\Delta v}{\Delta t}$ and what would the question wanting the maximum imply?

I see that delta V / delta T would be the acceleration and that the velocity would be greatest when acceleration is zero (maximum) but if I divide both sides by delta T I end up with:
-kx = m dV/dT
-kx = ma
Fn = ma
And I'm not really sure what I could do next...

 

[Student100]

I see that delta V / delta T would be the acceleration and that the velocity would be greatest when acceleration is zero (maximum) but if I divide both sides by delta T I end up with:
-kx = m dV/dT
-kx = ma
Fn = ma
And I'm not really sure what I could do next...

So you have $a=\frac{-kx}{m}$ or $a={w^2}{x}$, is this looking familiar?

 

[Thomas Brown]

I think I got it but can you please verify? I know that a = w^2 x is the acceleration in SHM, so I integrated the A(t) formula to get V(t) = xw * sin(wt) and velocity is greatest when sin(wt) = 1 so the maximum velocity is xw, or v = x * (k / m)^1/2.

 

[Student100]

I think I got it but can you please verify? I know that a = w^2 x is the acceleration in SHM, so I integrated the A(t) formula to get V(t) = xw * sin(wt) and velocity is greatest when sin(wt) = 1 so the maximum velocity is xw, or v = x * (k / m)^1/2.

Yes, $v(t)=- \omega xsin(\omega t)$ when you take the phase constant to be zero. So what you said originally was true, the maximum speed in this motion occurs when the acceleration is zero, as well as when the displacement from equilibrium is the same. (Also applies for the min!) $V_{max/min}= \pm \omega A$ depending on the direction of oscillation $V_{max}= \sqrt{\frac{k}{m}} x$ depending on your coordinate system choosen.

Is this the best way to show what you wanted to show? I can't think of another way from the impulse-momentum theorem off the top of my head, but looks good to me.

 

[Thomas Brown]

Yup, this makes sense! Thanks for your help!