A 200 gram block is attached to a spring with a spring constant of 8 N/m. The spring oscillates horizontally on a frictionless surface. Its velocity is 80 cm/s when x = - 4.2 cm.
a. What is the amplitude of oscillation?
b. What is the block’s maximum acceleration?
c. What is the block’s position when the acceleration is maximum?
d. What is the speed of the block when x = 2.5 cm?
I think I already got a and b but am confused on the other two parts
Acceleration is a maximum where the displacement from equilibrium (and thus the spring force) is a maximum.
Since KE + PE is a constant,
(1/2) k X^2 + (1/2) m V^2 = C
= 4*(0.042)^2 + (0.100)(0.080)^2
Note that I am using meters and not centimeters for x and V.
Once you know that constant, the total energy, you can compute BV at any value of x.
Okay I think I calculated that correctly but all the equations I see to use incorporate time which I do not have...I am sorry I am just really confused and my brain is about fried.
Nevermind, I'm dumb thanks!
The method that I used above is based upon conservation of energy and does not require you to know or calculate the time that x or V has a particular value. It is a useful shortcut.
"BV" in my previous post should be just V. Sloppy typing.
To solve parts c and d of this problem, we need to understand the relationship between position, velocity, acceleration, and time in simple harmonic motion. In simple harmonic motion, the position of an object is given by the equation:
x(t) = A * cos(ωt + φ)
where:
- x(t) is the position of the object at time t
- A is the amplitude of oscillation
- ω is the angular frequency (ω = 2πf, where f is the frequency)
- φ is the phase constant
The velocity and acceleration of the object can be derived by taking the derivatives of the position equation:
v(t) = -A * ω * sin(ωt + φ)
a(t) = -A * ω^2 * cos(ωt + φ)
Now let's solve parts c and d using the given information.
c. What is the block’s position when the acceleration is maximum?
To find the position when the acceleration is maximum, we need to determine the corresponding time for the maximum acceleration. In simple harmonic motion, the maximum acceleration occurs when the displacement is at its maximum or minimum.
Given that x = -4.2 cm when the velocity is 80 cm/s, we can use this information to find ω and φ.
Using the velocity equation, v(t) = -A * ω * sin(ωt + φ), and substituting the known values, we get:
80 cm/s = -A * ω * sin(ωt + φ)
We know that sin(ωt + φ) will be ±1 when the displacement is at its maximum or minimum, so we substitute sin(ωt + φ) = -1 into the equation above:
80 cm/s = A * ω
Since A = 4.2 cm and ω = 2πf, we can solve for f:
80 cm/s = 4.2 cm * 2πf
f = 80 cm/s / (4.2 cm * 2π) ≈ 3.8 Hz
Now we can find the angular frequency ω:
ω = 2πf ≈ 2π * 3.8 Hz ≈ 23.86 rad/s
With ω known, we can find φ:
Using the position equation, x(t) = A * cos(ωt + φ), and substituting t = 0 (since the acceleration is maximum when the object is at its extreme position), we get:
-4.2 cm = A * cos(φ)
cos(φ) = -4.2 cm / A ≈ -1
Since cos(φ) ≈ -1, φ ≈ π radians.
Now, in order to find the object's position when the acceleration is maximum, we substitute the known values into the position equation:
x(t) = A * cos(ωt + φ)
x(t) = 4.2 cm * cos(23.86 rad/s * t + π)
However, without the value of time (t), we cannot determine the exact position at this point. To find the position, we need additional information or need to know a specific time.
d. What is the speed of the block when x = 2.5 cm?
To find the speed of the block at a specific position, we need to determine the corresponding time for that position.
Using the given information that x = -4.2 cm when the velocity is 80 cm/s, we can use this to determine the phase constant φ.
Again, using the velocity equation, v(t) = -A * ω * sin(ωt + φ), and substituting the known values, we get:
80 cm/s = -A * ω * sin(ωt + φ)
We know that sin(ωt + φ) will be ±1 when the displacement is at its maximum or minimum, so we substitute sin(ωt + φ) = -1 into the equation above:
80 cm/s = A * ω
Since A = 4.2 cm and ω = 2πf, we can solve for f:
80 cm/s = 4.2 cm * 2πf
f = 80 cm/s / (4.2 cm * 2π) ≈ 3.8 Hz
Now we can find the angular frequency ω:
ω = 2πf ≈ 2π * 3.8 Hz ≈ 23.86 rad/s
With ω known, we can find φ:
Using the position equation, x(t) = A * cos(ωt + φ), and substituting the known value of x = 2.5 cm, we get:
2.5 cm = 4.2 cm * cos(23.86 rad/s * t + π)
To find the value of t, we rearrange the equation:
cos(23.86 rad/s * t + π) = 2.5 cm / 4.2 cm
Since the maximum value of cos is 1, we can determine that 2.5 cm / 4.2 cm should be between -1 and 1. Therefore, there are two possible values for the argument of the cosine function:
23.86 rad/s * t + π = arccos(2.5 cm / 4.2 cm)
or
23.86 rad/s * t + π = -arccos(2.5 cm / 4.2 cm)
Solving for t, we have:
t = (arccos(2.5 cm / 4.2 cm) - π) / 23.86 rad/s
or
t = (-arccos(2.5 cm / 4.2 cm) - π) / 23.86 rad/s
By substituting the value of t into the velocity equation, v(t) = -A * ω * sin(ωt + φ), we can determine the velocity of the block at that position.
Using the known values and the obtained values for t, we can find the answer.