The displacement of a block of mass 778g attached to a spring whose spring constant is 38N/m is given by x=Asin(ωt) where A=25cm. In the first complete cycle find the values of x and t at which the kinetic energy is equal to one half the potential energy.
First position:
First time:
Second position:
Second time:
Third position:
Third time:
Fourth position:
Fourth time:
time in s and position in cm please
forget cm, doing seconds and meters - SCI units
Pe = (1/2) k |x|^2
Ke = (1/2) m |v|^2
so for Ke = 1/2 Pe
m |v|^2 = .5 k |x|^2
but ω^2 = k/m
so
|v|^2 = .5 ω^2 |x|^2 At our spots
[A ω cos ωt]^2 = .5ω^2[A sin ωt]^2
A^2 ω^2 cos^2ωt = .5ω^2 A^2sin^2wt
cos^2 ωt = .5 sin^2 ωt
cos^2 ωt/sin^2 ωt = .5
cos/sin = + or - sqrt(1/2)
To find the values of x and t at which the kinetic energy is equal to one half the potential energy, we need to determine the positions and times during the first complete cycle when this condition is met.
Given:
Mass (m) = 778g = 0.778kg
Spring constant (k) = 38 N/m
Displacement equation: x = A * sin(ωt)
Amplitude (A) = 25cm = 0.25m
First, let's find the angular frequency (ω) using the formula: ω = sqrt(k/m)
ω = sqrt(38 N/m / 0.778 kg) ≈ 7.068 rad/s
To find the positions and times when the kinetic energy is equal to one-half the potential energy, we can set up the equation for kinetic energy and potential energy and solve for x and t.
1. First position:
Let's assume the initial position when t = 0 is the starting point of the first complete cycle. At this position, potential energy is maximum, and kinetic energy is zero.
Potential Energy (PE) = 1/2 * k * x^2
Kinetic Energy (KE) = 1/2 * m * (ω * A)^2
Set PE = 1/2 * KE
1/2 * k * x^2 = 1/2 * m * (ω * A)^2
1/2 * 38 N/m * x^2 = 1/2 * 0.778 kg * (7.068 rad/s * 0.25 m)^2
Solving this equation will give us the value of x when PE = KE/2.
2. Second position:
To find the second position where PE = KE/2, we can use the fact that potential energy is zero at the maximum displacement.
Potential Energy (PE) = 1/2 * k * x^2
Set PE = KE/2 and solve for x.
3. Third position:
To find the third position where PE = KE/2, we can repeat the calculation from step 2 with the opposite sign, as potential energy is zero at maximum displacement, whether positive or negative.
4. Fourth position:
Finally, to find the fourth position where PE = KE/2, we can use the fact that potential energy is zero at the starting position.
By solving the equations above, we can obtain the values of x and t at each required position.