Harmonic oscillation; new amplitude?

[6c 6f 76 65]

Homework Statement

A block with a mass M is located on a frictionless, horizontal surface and is attached to a horizontal spring with spring stiffness k. The block is being pulled out to the right a distance $x=x_0$ of equilibrium and released at $t = 0$.

At time $t_1$, corresponding to $\omega t_1=\varphi_1$, a lump of clay with mass m is dropped onto the block (sticking to it).

a) Use conservation of momentum, in the horizontal direction, to show that the new amplitude is:
$x_0\prime = x_0 \sqrt{\frac{M+m\cdot cos^2(\varphi_1)}{M+m}}$

b) Express the new angular frequency, $\omega\prime$ in terms of $\omega$

Homework Equations

$x(t)=x_0 \cdot cos(\omega\cdot t)$

$E=\frac{1}{2}kx_0^2$

$\omega=\sqrt{\frac{k}{m}}$

The Attempt at a Solution

a) Conservation of momentum:

$M\dot{x}=(M+m)\dot{x\prime}$

$M(-\omega\cdot sin(\omega\cdot t)\cdot x_0) =(M+m)\dot{x\prime}$, I'm not sure whether it should be $sin(\omega\cdot t_1)$ or $sin(\omega\cdot t)$

$-M\sqrt{\frac{k}{M}}\cdot sin(\omega\cdot t)\cdot x_0=(M+m)\dot{x\prime}$

$-\sqrt{M\cdot k}\cdot sin(\omega\cdot t)\cdot x_0=(M+m)\dot{x\prime}$, now I don't know what more to do. If I express $\dot{x\prime}$ using $sin(\omega \cdot t)$, ω and t will be different so i can't remove them later (unless i know the relationship between the angular frequencies, which is just what part b) is)

 

[vela]

To find the new amplitude, you need to know the total energy of the system after the collision. What's the kinetic energy immediately after the collision? What's the potential energy?

 

[6c 6f 76 65]

To find the new amplitude, you need to know the total energy of the system after the collision. What's the kinetic energy immediately after the collision? What's the potential energy?

Thanks for the help, here's what I got:
$M\dot{x}=(M+m)\dot{x}\prime$

$kx_0^2=kx^2\prime+(M+m)\dot{x}^2\prime$

$kx_0^2=kx^2\prime+\frac{(M+m)M^2\dot{x}^2}{(M+m)^2}$

$kx_0^2=kx^2\prime+\frac{(M+m)M^2\omega^2 x_0^2 sin^2(\omega t_1)}{(M+m)^2}$

$kx_0^2=kx^2\prime+\frac{M^2(\frac{k}{M}) x_0^2 sin^2(\omega t_1)}{(M+m)}$

$x_0^2=x^2\prime+\frac{M x_0^2 sin^2(\omega t_1)}{(M+m)}$

$x_0^2(1-\frac{M sin^2(\omega t_1)}{(M+m)})=x^2\prime$


$x_0^2(1-\frac{M (1-cos^2(\omega t_1))}{(M+m)})=x^2\prime$

$x_0^2(\frac{m+M}{m+M}-\frac{M (1-cos^2(\omega t_1))}{(M+m)})=x^2\prime$

$x_0^2(\frac{m+Mcos^2(\omega t_1)}{m+M})=x^2\prime$, not quite right $m+Mcos^2(\omega t_1)$ instead of $M+mcos^2(\omega t_1)$, and $x^2\prime \neq x_0^2\prime$

 

[vela]

Thanks for the help, here's what I got:
$M\dot{x}=(M+m)\dot{x}\prime$

Just to be clear, I assume you mean $M\dot{x}(t_1) = (M+m)\dot{x}'(t_1)$.

$kx_0^2=kx^2\prime+(M+m)\dot{x}^2\prime$

This isn't correct. You'd be saying that the original energy of the system, $\frac{1}{2}kx_0^2$, is the same as the energy after the lump is added. That's not the case because at $t=t_1$, the potential energy doesn't change, but the kinetic energy does.

 

[6c 6f 76 65]

That's not the case because at $t=t_1$, the potential energy doesn't change, but the kinetic energy does.

Ok, so $\frac{1}{2}M\dot{x(t_1)}=\frac{1}{2}(M+m)\dot{x\prime(t_1)}$? But if i substitute for either $\dot{x\prime(t_1)}$ or $\dot{x(t_1)}$ I end up with an equation of only one variable.

 

[vela]

No, this is an inelastic collision. Kinetic energy isn't conserved.

EDIT: Sorry, misread what you wrote. That's simply conservation of momentum again with a factor of 1/2 thrown in.

What's the potential energy right after the collision? What's the kinetic energy right after the collision?

 

[6c 6f 76 65]

What's the potential energy right after the collision? What's the kinetic energy right after the collision?

$PE=\frac{1}{2}k x\prime(t_1)^2$

$KE=\frac{1}{2}(M+m)\dot{x}\prime(t_1)^2$

 

[vela]

Right, so $E' = \frac{1}{2}kx_0'^2 = \frac{1}{2}kx'(t_1)^2 + \frac12(M+m)\dot{x}'(t_1)^2$. Now you just need to express $x_0'$ in terms of the unprimed variables.

 

[6c 6f 76 65]

Right, so $E' = \frac{1}{2}kx_0'^2 = \frac{1}{2}kx'(t_1)^2 + \frac12(M+m)\dot{x}'(t_1)^2$. Now you just need to express $x_0'$ in terms of the unprimed variables.

Thanks!

$k x_0'^2=\frac{M^2 \omega^2 x_0^2 sin^2(\omega t_1)}{M+m}+k x'^2(t_1)$

$k x_0'^2=\frac{M k x_0^2 sin^2(\omega t_1)}{M+m}+k x'^2(t_1)$, used $\omega^2=\frac{k}{M}$. $k$ is constant before and after the "collision".

$x_0'^2=\frac{M x_0^2 (1-cos^2(\omega t_1))}{M+m}+x'^2(t_1)$, any hints one the last prime term?

 

[haruspex]

$x_0'^2=\frac{M x_0^2 (1-cos^2(\omega t_1))}{M+m}+x'^2(t_1)$, any hints one the last prime term?

Isn't x'(t₁) the same as x(t₁)?

 

[6c 6f 76 65]

Isn't x'(t₁) the same as x(t₁)?

EDIT: Sorry, you're right! Thank you!