Oscillations and Waves of a spring

In summary: You should be able to work out the rest.In summary, the conversation is about a 2.00-kg object attached to a spring on a horizontal surface. The object is subjected to a force of 20.0 N and undergoes simple harmonic oscillations. The force constant of the spring is 1000 N/m, the frequency is 22.36 s^-1, and the maximum speed is 4.472 m/s. The maximum acceleration occurs when the object is at one third of its maximum displacement. The total energy of the system is equal to the elastic potential energy at any given point.
  • #1
ilovephysicssomuch
23
0
Can someone please help me with this question?:confused:

A 2.00-kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis) The object is now released from rest with an initial position of x= 0.200m, and it subsequenly undergoes simple harmonic oscillations. Find
(a) the force constant of the spring,
(b) the frequency of the oscillations, and
(c) the maximum speed of the object. Where does this maximum speed occur?
(d) Find the maximum acceleration of the object. Where does it occur?
(e) Find the total energy of the oscillation system.
(f) Find the speed
(g) Find the acceleration of the object when its position is equal to one third of the maximum value.
 
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  • #2
Can you show some of your working/thoughts?
 
  • #3
Tina is that you? Please show us your work lol.
 
  • #4
my work

Here is my work... I'm not 100% sure.. This questions is hard...
Mass = 2.00 kg
F= 20.0 N
x = 0.200m

(a) finding K
F= 0.5kx^2 if that is the right formula
20.0 N = 0.5k(0.2)^2
K= 1000 N/m

(b) Since w is also frequency but is in s^-1
w= (k/m)^1/2
w= (1000/2.00)^1/2
w= 22.36 s^-1

(c) Vmax = wA... What is A? I am treating A as being the same as x.. I don't know how to do this question.
Vmax = (22.36)(0.2)
Vmax = 4.472 m/s
Where does it occur?? no idea what formula to use.

(d) Amax = w^2A.. again?? is this the right formula?

(e) Total energy?
E = 0.5 kA^2 or is it kinetic energy + potential energy??

(f) Find the speed.. formula for speed is
v = w/k..
so if this is right.. then
22.36/1000.. so v is 0.02236

(g) No idea


Please Help me!:confused:
 
  • #5
(a) The spring constant, [itex]k[/itex] is given by Hooke's law for springs: [tex]F = -kx [/tex]. I think you answer is wrong by a factor of ten.

(b)It looks like you've worked out [itex]\omega[/itex] correctly but remember [tex]f = \frac{\omega}{2\pi}[/tex]. Your error in (a) follows through here.

(c)Your formula is correct but there are follow through errors from the previous questions, A is amplitude or maximum displacement. As the spring is stretch it gains elastic potential energy. Once the spring is released and begins to recoil the elastic potential energy is converted to kinetic energy. When the spring is in neither compression or tension there is no elastic potential energy in the spring, the energy of the system must remain constant therefore all of the elastic potential energy must be in the form of kinetic energy. So where is the velocity the greatest?

(d)Correct formula except [itex] a = -\omega^2 x [/itex] because the acceleration is always in the opposite direction to the displacement (towards the equilibrium position)

(e)Your forumla [itex]E = \frac{1}{2} k A [/itex]is correct.

(f)I don't understand the question. Find the speed when? The speed is variable in any mass/spring system.

(g)[tex] a = -\omega^2 x [/tex] where x is displacement. You know the amplitude so [itex] x = \frac{1}{3}A [/itex]
 

FAQ: Oscillations and Waves of a spring

What is the definition of oscillation?

Oscillation is the repetitive back and forth motion of an object around a central point or equilibrium position.

How does a spring oscillate?

When a spring is displaced from its equilibrium position, it exerts a restoring force that pulls it back towards the equilibrium point. This causes the spring to oscillate back and forth until eventually coming to rest at the equilibrium position.

What factors affect the frequency of oscillation in a spring?

The frequency of oscillation in a spring is affected by the mass of the object attached to the spring, the spring constant (a measure of its stiffness), and the amplitude of the oscillation (maximum displacement from equilibrium).

What is the relationship between the period and frequency of a spring's oscillation?

The period (T) of a spring's oscillation is the time it takes to complete one full cycle of motion, while the frequency (f) is the number of cycles per unit time. The period and frequency are inversely related, meaning that as one increases, the other decreases. This relationship can be represented by the equation f = 1/T.

How does the amplitude of a spring's oscillation affect its energy?

The amplitude of a spring's oscillation directly affects its energy. As the amplitude increases, so does the energy of the oscillation. This is because the energy of a spring is proportional to the square of its amplitude. So, a larger amplitude results in a larger energy and a smaller amplitude results in a smaller energy.

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