A mass m is attached to a spring of constant k. The spring oscillates in simple harmonic motion (amplitude A). In terms of A:
a) At what displacement from equilibrium is the speed half of the max value and b) at what displacement from equilibrium is the potential energy half of the max value? Why is letter b greater than letter a?
Any help is appreciated.
Say x = A sin wt
then
v = A w cos w t
max speed is when cos w t = 1 or -1
That is when sin w t = 0
which is when x = 0
1/2 max is when cos w t = 1/2
We need to know sin wt when cos w t = 1/2
BUT we know that cos^2+sin^2 = 1 (trig)
so
(1/2)(1/2) + sin^2 wt = 1
sin^2 wt = 3/4
sin wt = .866
so
x = .866 A at half max speed
-------------------------
Now PE = (1/2) k x^2
max PE = .5 k A^2
when is PE = (1/2)(1/2) k A^2 ???
(1/2) k x^2 = (1/2)(1/2) k A^2
x^2 = .5 A^2
x = .707 A
Ah, I really appreciate that. Thank you.
The kinetic energy + potential energy is constant. Each goes to zero when the other is max.
Therefore when the PE is 1/2 max, the KE will also be (1/2) max
the half PE occurs when x = .707 A, part b
so the half KE has to happen at x = .707 A as well
that means (1/2) m v^2 = (1/2)(1/2)m Vmax^2
at x = .707 A
v^2 = .5 Vmax^2 at x = .707 A
so
v = .707 Vmax at .707 A
THEREFORE v does not reach .5 Vmax until further out than .707 A (speed max at x = 0, and speed 0 at x = A)
SO - I claim that the answer to b is SMALLER than the answer to a
To solve this problem, we need to use the equations of motion for simple harmonic motion. The displacement of the mass from equilibrium at time t is given by the equation:
x(t) = A * cos(ωt)
where x(t) is the displacement at time t, A is the amplitude of the motion, and ω is the angular frequency given by ω = √(k/m), with k being the spring constant and m being the mass.
The velocity of the mass at time t is given by the derivative of the displacement equation:
v(t) = -A * ω * sin(ωt)
To find the displacement from equilibrium at which the speed (magnitude of velocity) is half of the maximum value, we need to solve the following equation:
|v(t)| = 0.5 * |v_max|
where |v_max| = A * ω is the maximum speed.
Substituting the expression for v(t), we have:
0.5 * |v_max| = 0.5 * A * ω = |A * ω * sin(ωt)|
Dividing both sides of the equation by |A * ω|, we get:
0.5 = |sin(ωt)|
Since the sine function varies between -1 and 1, this equation is satisfied when:
sin(ωt) = ±0.5
To find the corresponding displacement, we substitute this result into the displacement equation:
x(t) = A * cos(ωt)
For sin(ωt) = ±0.5, we have:
cos(ωt) = ±√(1 - sin^2(ωt)) = ±√(1 - 0.5^2) = ±√(1 - 0.25) = ±√(0.75)
This means that when the speed is half of the maximum value, the displacement from equilibrium is given by:
x(t) = ±A * √(0.75)
To find the displacement from equilibrium at which the potential energy is half of the maximum value, we use the equation for potential energy in simple harmonic motion, which is given by:
U(t) = 0.5 * k * x^2(t)
where U(t) is the potential energy at time t.
We want to solve the following equation:
U(t) = 0.5 * U_max
Substituting the expression for U(t), we have:
0.5 * k * x^2(t) = 0.5 * (0.5 * k * A^2)
Cancelling the factors of 0.5 and k, we get:
x^2(t) = 0.5 * A^2
Taking the square root of both sides, we have:
x(t) = ±√(0.5) * A
Comparing the expressions for x(t) in question a) and question b), we can see that the displacement from equilibrium at which the potential energy is half of the maximum value (question b) is greater than the displacement from equilibrium at which the speed is half of the maximum value (question a). This is because the potential energy is proportional to the square of the displacement, while the speed (magnitude of velocity) is only determined by the displacement. Hence, for the same factor (half), the displacement required to achieve half of the maximum potential energy is greater than the displacement required to achieve half of the maximum speed.