Determine the frequency of the resulting motion

In summary, a 1.2-kg block oscillates with an amplitude of 10 cm and a phase constant of phi=-pi/2 when it is struck by a mass of 0.80 kg moving from the right at 1.7 m/s.
  • #1
akan
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Homework Statement


A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 27 N/m. The block is oscillating with amplitude 10 cm and phase constant phi = -pi/2. A block of mass 0.80 kg is moving from the right at 1.7 m/s. It strikes the first block when the latter is at the rightmost point in its oscillation. The collision is completely inelastic, and the two blocks stick together.

a) Determine the frequency of the resulting motion.
I have found this to be 0.58 Hz, and this is the answer.

b) Determine the amplitude of the resulting motion.
E = (1/2)kx^2 + (1/2)mv^2 = 1/2k(A_new)^2 (I think?)
kx^2 + mv^2 = k(A_new)^2
(27)(.1)^2 + (.8)(1.7)^2 = (27)(A_new)^2
A_new = .332

This is what I get, but it does not seem right, and is not accepted as a valid answer... If my approach is wrong, then how do I solve this part?

c) Determine the phase constant (relative to the original t = 0) of the resulting motion.

Well, the relevant equations, I suppose, are:

x = A cos(wt + phi)
v = -w A sin (wt + phi)
a = -w A cos (wt + phi)

However, even once I have the amplitude, I don't know how to solve this part of the question. Please give a hint or something. Thanks.
 
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  • #2
Hi akan,

akan said:

Homework Statement


A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 27 N/m. The block is oscillating with amplitude 10 cm and phase constant phi = -pi/2. A block of mass 0.80 kg is moving from the right at 1.7 m/s. It strikes the first block when the latter is at the rightmost point in its oscillation. The collision is completely inelastic, and the two blocks stick together.

a) Determine the frequency of the resulting motion.
I have found this to be 0.58 Hz, and this is the answer.

b) Determine the amplitude of the resulting motion.
E = (1/2)kx^2 + (1/2)mv^2 = 1/2k(A_new)^2 (I think?)
kx^2 + mv^2 = k(A_new)^2
(27)(.1)^2 + (.8)(1.7)^2 = (27)(A_new)^2
A_new = .332

When they refer to the resulting motion, they mean the motion after the collision. Immediately after the collision, what is the total mass on the spring, and what speed does it have? (You have 0.8kg and 1.7m/s, but that is the mass of the moving block and its speed just before the collision.) Once you correct that, I believe you will get the right answer.
 
  • #3
m1v1 + m2v1 = v2(m1+m2).
m2v1 = v2(m1+m2)
v2 = m2v1 / (m1+m2)

And this is the velocity I use in the equation, right? Is everything else correct? Thanks.
 
  • #4
akan said:
m1v1 + m2v1 = v2(m1+m2).
m2v1 = v2(m1+m2)
v2 = m2v1 / (m1+m2)

And this is the velocity I use in the equation, right? Is everything else correct? Thanks.

That looks like the right idea. Use that velocity and the total mass.
 
  • #5
That gave me the correct result. However, how do I solve part C? I guess what I don't undersand is what the wt term, and what do I substitute in there. I understand that t is a point in time, but what's it equal to? w is an angular velocity, but what is that equal to? The angular velocity of the system or at the point of time? Thanks.
 
Last edited:
  • #6
Ok, so x = A cos (wt + phi)
If we take the new t = 0 as the moment of collision, phi = arccos(x/A) = arccos((.10)/.21) = 58.41186449.
But relative to the original t = 0, which had a constant of -pi/2 = -90 degrees, we have it that the new phi = 58.41186449 - 90 = -28.44. But that's not accepted as a valid answer. Any hints?
 
Last edited:

1. What is frequency in relation to motion?

In physics, frequency is defined as the number of cycles or oscillations per unit of time. In the context of motion, frequency refers to the number of times an object moves back and forth within a given time frame.

2. How is frequency calculated for motion?

Frequency is calculated by dividing the number of cycles or oscillations by the time it takes to complete one cycle. This can be represented by the equation frequency = cycles/time.

3. Does the frequency of motion affect its speed?

No, the frequency of motion does not directly affect its speed. The speed of an object in motion is determined by its velocity, which is the rate of change in its position over time.

4. What factors can affect the frequency of motion?

The frequency of motion can be affected by the amplitude (size) of the oscillations, the mass of the object, and any external forces acting on the object.

5. How is the frequency of motion measured?

The frequency of motion can be measured using a tool called a frequency counter, which counts the number of cycles or oscillations in a given time frame. It can also be determined by measuring the period of the motion (time for one cycle) and using the equation frequency = 1/period.

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