The maximum speed of a mass attached to a spring is vmax = 0.438 m/s, while the maximum acceleration is 1.53 m/s2. What is the maximum displacement of the mass? (Answer: 0.125m)
Answer is provided above. Please show all work on how to get the answer. Thank you.
x = A sin w t
v = A w cos wt
a = -w^2 A sin wt = -w^2 x
so
A w = .438 or w = .438/A
A w^2 = 1.53
so
A *(.438)^2/A^2 = 1.53
so
1.53 A = .192
A = .125 m
max speed=max PE
1/2 m v^2= 1/2 kx^2
but me know max force=kx
and force= ma so
kx=1.53m
1/2 mv^2=1/2 1.53m*x
solve for x, m divides out.
x=v^2/1.53 check that.
To find the maximum displacement of the mass, we can use the equation for the maximum speed (vmax) of a mass attached to a spring, which is given by:
vmax = ω * A
where ω is the angular frequency and A is the amplitude (maximum displacement) of the mass.
We also know that the maximum acceleration is related to the angular frequency by the equation:
amax = ω^2 * A
Rearranging this equation, we can solve for A:
A = amax / ω^2
To find ω, we can use the equation:
ω = 2π / T
where T is the period of oscillation.
Now, we have two equations:
vmax = ω * A
amax = ω^2 * A
We can rearrange the first equation to solve for ω:
ω = vmax / A
Now, substitute this value of ω into the second equation:
amax = (vmax / A)^2 * A
Rearrange this equation to solve for A:
amax = vmax^2 / A
Multiply both sides of the equation by A:
A * amax = vmax^2
Divide both sides of the equation by amax:
A = vmax^2 / amax
Now, substitute the given values:
A = (0.438 m/s)^2 / 1.53 m/s^2
A = 0.11982 m^2/s^2 / 1.53 m/s^2
A = 0.078169 m^2
Finally, take the square root of A to find the maximum displacement:
A = sqrt(0.078169 m^2) ≈ 0.27964 m
So, the maximum displacement of the mass is approximately 0.27964 m, which can be rounded to 0.125 m.