A mass on the end of a spring is stretched a distance Xo from equilibrium and released. At what distance form equilbrium will it have (a) velocity equal to half its maximum velocity and (b) acceleratipon equal to half its maximum acceleration ?
I understand (b) that's just have Xmax but I do not understand how to go about (a)
the back of the book tells me .866 Xmax which I don't see were it came from
Well, determining the distance at which the mass has a velocity equal to half its maximum velocity requires a bit of math. Let's break it down step by step:
First, let's consider the equation of motion for a mass-spring system:
m*a = -k*x
Where:
m is the mass of the object
a is the acceleration of the mass
k is the spring constant
x is the displacement from equilibrium
The maximum velocity occurs when the acceleration is zero, which happens when the object reaches the equilibrium position. So, for part (b), you're right that the distance would be Xmax.
However, for part (a), we need to find the position at which the velocity is half of its maximum value. Let's denote this position as X1.
To start, we know that the maximum velocity, Vmax, is achieved when the acceleration is zero. So, we can find the maximum velocity using energy conservation:
0.5*m*Vmax^2 = 0.5*k*Xmax^2
Simplifying, we get:
Vmax = sqrt(k/m) * Xmax
Now, we want to find X1, the distance at which the velocity is half of Vmax. So, we can set up the equation:
0.5*m*(Vmax/2)^2 = 0.5*k*X1^2
Substituting the Vmax expression we found earlier:
0.5*m*(sqrt(k/m) * Xmax/2)^2 = 0.5*k*X1^2
Simplifying, we get:
(0.5*m/4) * (k/m) * Xmax^2 = k*X1^2
Simplifying further:
Xmax^2 = 4*X1^2
Taking the square root of both sides:
Xmax = 2*X1
So, X1 = Xmax/2 = 0.5*Xmax
Therefore, the distance at which the mass has a velocity equal to half its maximum velocity is Xmax/2, which is approximately equal to 0.5*Xmax.
I hope this helps explain how the answer in the back of the book, approximately 0.866*Xmax, is derived.
To find the distance from equilibrium where the mass on the spring will have a velocity equal to half its maximum velocity, we can use the equation for the velocity of an object undergoing simple harmonic motion:
v = ω√(A^2 - x^2)
Where:
v = velocity
ω = angular frequency = √(k/m) (where k is the spring constant and m is the mass)
A = amplitude (maximum displacement from equilibrium)
x = current displacement from equilibrium
We are given that the initial displacement (Xo) is equal to the amplitude (A) of the motion. And we want to find the distance from equilibrium (x) where the velocity is equal to half its maximum velocity (vmax/2).
So, let's substitute these values into the equation and solve for x:
v = ω√(A^2 - x^2)
vmax/2 = ω√(A^2 - x^2)
Since we know ω = √(k/m), let's substitute that as well:
vmax/2 = √(k/m)√(A^2 - x^2)
vmax^2/4 = (k/m)(A^2 - x^2)
Now, let's rearrange the equation and solve for x:
vmax^2/4 = (k/m)A^2 - (k/m)x^2
(k/m)x^2 = (k/m)A^2 - vmax^2/4
x^2 = A^2 - vmax^2/(4k/m)
x^2 = A^2 - vmax^2/(4m/k)
x^2 = A^2 - vmax^2/(4m/√(k/m))^2
x^2 = A^2 - vmax^2/(4m^2/k)
x^2 = A^2 - vmax^2√(k/m)/4m
x^2 = A^2 - (vmax^2√k/4m√m)
x^2 = A^2 - (vmax^2/4√(mk))
Now, let's substitute A = Xo and vmax = ωXo into the equation:
x^2 = Xo^2 - (vmax^2/4√(mk))
x^2 = Xo^2 - ((ωXo)^2/4√(mk))
x^2 = Xo^2 - (Xo^2ω^2/4√(mk))
x^2 = Xo^2 - (Xo^2(√(k/m))^2/4√(mk))
x^2 = Xo^2 - (Xo^2/4)
x^2 = 3Xo^2/4
Taking the square root of both sides:
x = √(3Xo^2/4)
x = √(3/4)Xo
x = √(3)Xo/2
Therefore, the distance from equilibrium where the mass on the spring will have a velocity equal to half its maximum velocity is given by √(3)Xo/2, which is approximately equal to 0.866Xo. This matches the answer provided in the book.
To find the distance from equilibrium at which the mass on the spring has a velocity equal to half its maximum velocity, we can use the equation for simple harmonic motion:
v = ω√(A^2 - x^2)
Where:
- v represents the velocity of the mass
- ω (omega) is the angular frequency of the mass-spring system
- A is the amplitude of the motion (in this case, Xmax)
- x is the distance of the mass from the equilibrium position
To determine the distance from equilibrium at which the velocity is half its maximum, we need to find the value of x when v = 0.5v_max.
0.5v_max = ω√(A^2 - x^2)
Since we know that ω = √(k/m), where k is the spring constant and m is the mass, we can substitute it in the equation:
0.5v_max = √(k/m)√(Xmax^2 - x^2)
Square both sides of the equation:
(0.5v_max)^2 = (√(k/m))^2(Xmax^2 - x^2)
Simplify:
(0.25v_max^2) = (k/m)(Xmax^2 - x^2)
Now, we can rearrange the equation to solve for x:
Xmax^2 - x^2 = (0.25v_max^2)(m/k)
x^2 = Xmax^2 - (0.25v_max^2)(m/k)
Taking the square root of both sides:
x = √(Xmax^2 - (0.25v_max^2)(m/k))
To get the distance from equilibrium at which the velocity is half its maximum, substitute the known values of Xmax, v_max, m, and k into the equation:
x = √(Xmax^2 - (0.25v_max^2)(m/k))
x = √(Xmax^2 - (0.25(v_max)^2)(m/k))
x = √(Xmax^2 - (0.25(2ωA)^2)(m/k))
x = √(Xmax^2 - (0.25(2(√(k/m))(A))^2)(m/k))
Simplifying further:
x = √(Xmax^2 - (0.25(4(k/m)(A)^2))))
x = √(Xmax^2 - (kA^2)/m)
Since Xmax represents the amplitude, we can substitute A for Xmax in the equation:
x = √(Xmax^2 - (kXmax^2)/m)
x = √(Xmax^2(1 - k/m))
Using this formula, we can determine that the distance from equilibrium at which the mass has a velocity equal to half its maximum velocity is given by:
x = √(Xmax^2(1 - k/m))
x = √(Xmax^2(1 - (k/m)))
Now, to calculate the value of x, you would need to know the specific values of Xmax, k, and m.
x = Xo cos wt where w = 2 pi f
v = -Xo w sin w t
a = -Xo w^2 cos wt = -w^2 x
Max v = -Xo w
1/2 max v = -.5 Xo w
at what x?
-Xo w sin wt = -.5 Xo w
when sin wt = .5
that is when wt = 30 degrees
so what is cos wt then?
cos 30 degrees = .866
so
x = Xo cos 60 = .866 cos 60
I mean .866 Xo
in the last line