a vertical spring with spring stiffness constant 305 n/m oscillates with an amplitude of 28.0 cm when 0.260 kg hangs from it. the mass passes through the equilibrium point (y=0) with positive velocity at t=0. (a) what equation describes this motion as a function of time? (b) at what times will the spring be longest and shortest?
To answer this question, we can use the equation that describes the motion of a mass-spring system:
m * (d^2y/dt^2) + k * y = 0
Where:
- m is the mass (0.260 kg in this case)
- y is the displacement from the equilibrium point (in meters)
- t is the time (in seconds)
- k is the spring stiffness constant (305 N/m in this case)
Now, let's solve this equation and find the equation that describes the motion as a function of time (a), and determine the times when the spring is longest and shortest (b).
(a) Firstly, let's assume that the general solution to this equation is of the form y(t) = A * cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle.
Now, let's find the values of ω and A.
Given:
- Amplitude, A = 0.28 m (converted from 28.0 cm)
- m = 0.260 kg
- k = 305 N/m
We can identify that the angular frequency ω is related to the mass and the spring constant using the equation ω = √(k / m).
Substituting the values:
ω = √(305 N/m / 0.260 kg) ≈ 9.72 rad/s
So, the equation describing the motion is:
y(t) = 0.28 * cos(9.72t + φ)
(b) To find the times when the spring is longest and shortest, we need to determine the maximum and minimum values of y(t).
Since the amplitude A is the maximum displacement of the mass from the equilibrium point, the longest and shortest states occur when y(t) = ± A. Therefore, we have:
When y(t) = A:
0.28 * cos(9.72t + φ) = 0.28
When y(t) = -A:
0.28 * cos(9.72t + φ) = -0.28
We can solve these equations for t to find the times when the spring is longest and shortest.
For y(t) = A:
cos(9.72t + φ) = 1
For y(t) = -A:
cos(9.72t + φ) = -1
In both cases, we know that the cosine function equals 1 or -1 at certain points in its cycle. So, we can set ωt + φ to the corresponding values.
For y(t) = A:
9.72t + φ = 0
For y(t) = -A:
9.72t + φ = π
Solving these equations, we can find the values of t when the spring is longest and shortest.
Keep in mind that the values of φ may cancel out in these equations due to the periodic nature of the cosine function. Hence, the absolute phase angle is not essential for this calculation.
To describe the motion of the mass hanging from the vertical spring as a function of time, we need to consider the equation for simple harmonic motion (SHM). In this case, the mass hanging from the spring causes the spring to elongate and contract, resulting in oscillatory motion.
(a) Equation for Simple Harmonic Motion:
The equation for SHM is given by:
y(t) = A * cos(ωt + φ)
Where:
y(t) = Displacement of the mass from the equilibrium position at time t
A = Amplitude of oscillation (maximum displacement)
ω = Angular frequency
t = Time
φ = Phase constant
In this case, the amplitude of oscillation is given as 28.0 cm, which is equal to 0.28 m.
To determine the angular frequency (ω), we can use the formula:
ω = sqrt(k / m)
Where:
k = Spring stiffness constant = 305 N/m
m = Mass hanging from the spring = 0.260 kg
Substituting the given values:
ω = sqrt(305 N/m / 0.260 kg)
≈ 7.8047 rad/s (rounded to 5 decimal places)
Now, the equation for the motion of the mass hanging from the vertical spring is:
y(t) = 0.28 * cos(7.8047t + φ)
(b) Length of the spring:
The length of the spring (L) is given by the equilibrium position plus the displacement of the mass:
L = y(t) + 0 (since the equilibrium position is y = 0)
The spring will be longest when the mass reaches its maximum displacement. At this point, y(t) is equal to the amplitude of oscillation (A). Therefore, the spring will be longest at t = 0.
The spring will be shortest when the mass is at the equilibrium position, where y(t) is equal to 0. This occurs twice during each oscillation. So, the spring will be shortest at t = T/2 and t = 3T/2, where T is the time period of the oscillation.
The time period of the oscillation (T) can be calculated using the formula:
T = 2π / ω
Substituting the value of ω calculated earlier:
T = 2π / 7.8047 rad/s
≈ 0.8055 s (rounded to 4 decimal places)
Therefore, the spring will be shortest at t = (0.8055 s)/2 and t = (0.8055 s)*(3/2), i.e., t = 0.4027 s and t = 1.208 s.
(a)
ω=sqrt(k/m) = sqrt(305/0.26) = 34.2 rad/s
y=Asin ωt = >
y=0.28sin34.2t (meters)
(b) y=A (longest)
A= Asin ωt
sin ωt = 1
ωt= π
t= π/ω=3.14/34.2= 0.092 s.
T=2π/ ω = 2•3.14/34.2 = 0.183 s
The shortest spring after T/2 =>
t₁ =t+ (T/2) = 0.092 +(0.183/2) =
=0.183 s