A mass is oscillating with amplitude A at the end of a spring.
How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?
x=A•sinωt
v =dx/dt = A•ω•cosωt
PE = KE
kx^2/2 = mv^2/2
k = m•ω^2
m•ω^2• (A•sinωt)^2/2 = m•( A•ω•cosωt)^2/2
(sinωt)^2 = (cosωt)^2
(tan ωt)^2 = 1
ωt = π/4
x=A•sinωt = A•sin(π/4) =0.707•A
It is not clear
Well, brace yourself for a physics joke!
When the elastic potential energy equals the kinetic energy, it means the mass has reached "equi-librium"! So, to answer your question, the mass is exactly A-funny distance away from the equilibrium position of the spring. Gotta love those balanced physics puns, don't you?
When the elastic potential energy equals the kinetic energy, the displacement (x) of the mass from its equilibrium position can be determined.
First, let's consider the equations for kinetic energy (KE) and elastic potential energy (PE) in terms of the displacement of the mass from equilibrium.
The kinetic energy of the mass is given by:
KE = (1/2)mv^2
Where:
m is the mass of the object
v is the velocity of the mass
The elastic potential energy of the spring is given by:
PE = (1/2)kx^2
Where:
k is the spring constant
x is the displacement of the mass from its equilibrium position
Given that PE = KE, we can set these two equations equal to each other:
(1/2)kx^2 = (1/2)mv^2
Canceling out the common factors, we have:
kx^2 = mv^2
Now, we can solve for x:
x^2 = (mv^2) / k
Taking the square root of both sides, we find:
x = √(mv^2 / k)
Since we're looking for the displacement in terms of the amplitude (A), we can express m, v, and k in terms of A:
m = mass of the object
v = maximum velocity of the object, which can be expressed as v = ωA, where ω is the angular frequency
k = spring constant
Substituting these values, we have:
x = √(((mA(ωA)^2) / k)
= A√((mA(ωA)^2) / k)
= A√((m(A^2 ω^2)) / k)
= A√(mA^2 ω^2 / k)
= A√(mA^2 ω^2 / k)
= A√(A^2 ω^2 m / k)
Simplifying further, we have:
x = A√(m/k)
So, the displacement of the mass from the equilibrium position when the elastic potential energy equals the kinetic energy is given by x = A√(m/k).
To determine how far the mass is from the equilibrium position when the elastic potential energy equals the kinetic energy, we need to equate the two energy expressions and solve for the displacement.
Let's start by expressing the elastic potential energy, which is given by:
Elastic Potential Energy (PE) = (1/2) k x^2
Here, k is the spring constant and x represents the displacement of the mass from the equilibrium position.
Next, let's consider the kinetic energy, given by:
Kinetic Energy (KE) = (1/2) m v^2
Here, m represents the mass of the object, and v is the velocity of the mass at any given point during the oscillation.
Knowing that the total mechanical energy (E) remains constant, we can equate the two energy expressions:
PE = KE
(1/2) k x^2 = (1/2) m v^2
Now, we can use the relationship between velocity and displacement for an object undergoing simple harmonic motion. The velocity as a function of displacement is given by:
v = A ω sin(ωt + φ)
In this equation, A represents the amplitude of oscillation, ω is the angular frequency (ω = sqrt(k/m)), t is the time, and φ is the phase angle.
Substituting this expression for velocity into our earlier equation, we have:
(1/2) k x^2 = (1/2) m (A ω sin(ωt + φ))^2
Simplifying further:
(1/2) k x^2 = (1/2) m A^2 ω^2 sin^2(ωt + φ)
Dividing both sides by (1/2) m A^2 ω^2:
x^2 = sin^2(ωt + φ)
Finally, we solve for x (the displacement) in terms of A (the amplitude) when the elastic potential energy equals the kinetic energy. Taking the square root:
x = ± A sin(ωt + φ)
Therefore, the mass is at a distance of A from the equilibrium position when the elastic potential energy equals the kinetic energy.