Harmonic oscillation; new amplitude?

In summary, using conservation of momentum and energy, the new amplitude for a block attached to a horizontal spring with a lump of clay dropped on it is given by x_0\prime = x_0 \sqrt{\frac{M+m\cdot cos^2(\varphi_1)}{M+m}}. The new angular frequency, \omega\prime, can be expressed in terms of the original angular frequency, \omega, as \omega\prime=\sqrt{\frac{k}{M+m}}.
  • #1
6c 6f 76 65
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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 [itex]x=x_0[/itex] of equilibrium and released at [itex]t = 0[/itex].

At time [itex]t_1[/itex], corresponding to [itex]\omega t_1=\varphi_1[/itex], 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:
[itex]x_0\prime = x_0 \sqrt{\frac{M+m\cdot cos^2(\varphi_1)}{M+m}}[/itex]

b) Express the new angular frequency, [itex]\omega\prime[/itex] in terms of [itex]\omega[/itex]

Homework Equations


[itex]x(t)=x_0 \cdot cos(\omega\cdot t)[/itex]

[itex]E=\frac{1}{2}kx_0^2[/itex]

[itex]\omega=\sqrt{\frac{k}{m}}[/itex]

The Attempt at a Solution


a) Conservation of momentum:

[itex]M\dot{x}=(M+m)\dot{x\prime}[/itex]

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

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

[itex]-\sqrt{M\cdot k}\cdot sin(\omega\cdot t)\cdot x_0=(M+m)\dot{x\prime}[/itex], now I don't know what more to do. If I express [itex]\dot{x\prime}[/itex] using [itex]sin(\omega \cdot t)[/itex], 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)
 
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  • #2
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?
 
  • #3
vela said:
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:
[itex]M\dot{x}=(M+m)\dot{x}\prime[/itex]

[itex]kx_0^2=kx^2\prime+(M+m)\dot{x}^2\prime[/itex]

[itex]kx_0^2=kx^2\prime+\frac{(M+m)M^2\dot{x}^2}{(M+m)^2}[/itex]

[itex]kx_0^2=kx^2\prime+\frac{(M+m)M^2\omega^2 x_0^2 sin^2(\omega t_1)}{(M+m)^2}[/itex]

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

[itex]x_0^2=x^2\prime+\frac{M x_0^2 sin^2(\omega t_1)}{(M+m)}[/itex]

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


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

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

[itex]x_0^2(\frac{m+Mcos^2(\omega t_1)}{m+M})=x^2\prime[/itex], not quite right [itex]m+Mcos^2(\omega t_1)[/itex] instead of [itex]M+mcos^2(\omega t_1)[/itex], and [itex]x^2\prime \neq x_0^2\prime[/itex]
 
  • #4
6c 6f 76 65 said:
Thanks for the help, here's what I got:
[itex]M\dot{x}=(M+m)\dot{x}\prime[/itex]
Just to be clear, I assume you mean ##M\dot{x}(t_1) = (M+m)\dot{x}'(t_1)##.

[itex]kx_0^2=kx^2\prime+(M+m)\dot{x}^2\prime[/itex]
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.
 
  • #5
vela said:
That's not the case because at ##t=t_1##, the potential energy doesn't change, but the kinetic energy does.

Ok, so [itex]\frac{1}{2}M\dot{x(t_1)}=\frac{1}{2}(M+m)\dot{x\prime(t_1)}[/itex]? But if i substitute for either [itex]\dot{x\prime(t_1)}[/itex] or [itex]\dot{x(t_1)}[/itex] I end up with an equation of only one variable.
 
  • #6
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?
 
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  • #7
vela said:
What's the potential energy right after the collision? What's the kinetic energy right after the collision?

[itex]PE=\frac{1}{2}k x\prime(t_1)^2[/itex]

[itex]KE=\frac{1}{2}(M+m)\dot{x}\prime(t_1)^2[/itex]
 
  • #8
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.
 
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  • #9
vela said:
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!

[itex]k x_0'^2=\frac{M^2 \omega^2 x_0^2 sin^2(\omega t_1)}{M+m}+k x'^2(t_1)[/itex]

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

[itex]x_0'^2=\frac{M x_0^2 (1-cos^2(\omega t_1))}{M+m}+x'^2(t_1)[/itex], any hints one the last prime term?
 
  • #10
6c 6f 76 65 said:
[itex]x_0'^2=\frac{M x_0^2 (1-cos^2(\omega t_1))}{M+m}+x'^2(t_1)[/itex], any hints one the last prime term?
Isn't x'(t1) the same as x(t1)?
 
  • #11
haruspex said:
Isn't x'(t1) the same as x(t1)?

EDIT: Sorry, you're right! Thank you!
 
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Related to Harmonic oscillation; new amplitude?

What is harmonic oscillation?

Harmonic oscillation is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. This means that the object or system will continue to oscillate back and forth between two points at a constant frequency.

What is the formula for harmonic oscillation?

The formula for harmonic oscillation is x = A*sin(ωt + φ), where x is the displacement, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

What is the new amplitude in harmonic oscillation?

The new amplitude in harmonic oscillation refers to a change in the maximum displacement of the object or system from its equilibrium position. This can occur due to a change in the frequency or energy of the oscillation.

What factors affect harmonic oscillation?

The factors that affect harmonic oscillation include the mass of the object or system, the spring constant, and the amplitude of the oscillation. Additionally, any external forces or damping can also affect the oscillation.

What are some real-life examples of harmonic oscillation?

Some real-life examples of harmonic oscillation include a swinging pendulum, a bouncing spring, and the vibrations of a guitar string. In each of these cases, the system oscillates back and forth in a regular pattern at a constant frequency.

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