A 1.25kg mass stretches a vertical spring 0.215m.If the spring is stretches an additional 0.130m and released,
(a)how long does it take to reach the equilibrium position again?
(b)Write an equation giving its position y as a function of time t.
(c)What will be its maximum speed and maximum acceleration?
The spring constant is
k = Mg/0.215m = 57.0 N/m
The frequency of oscillation is
f = [1/(2 pi)]sqrt(k/m) = 1.075 Hz
The angular frequency is
w = sqrt(k/m) = 6.75 rad/s
The period is 1/f = 0.93 seconds
(a) 1/4 period = 0.2325 s is the time needed to return to the equilibrium position
(b) If y is measured from the equilibrium position
y = 0.13 cos wt = 0.14 cos6.75t
(c) Max speed = w*(amplitude)= 0.13 w
Max acceleration = w^2*(amplitude)
(a) Oh, well, the time it takes to reach equilibrium again is directly proportional to how fast the spring rebounds, or in other words, how determined it is! So, we need to look at the spring constant. Do you know it? Otherwise, we'll have to find it using the Hooke's law.
(b) To express the position y as a function of time t, we can use simple harmonic motion equations. Let me put on my mathematician hat for a moment. Ahem, here's the equation: y(t) = A*cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. To determine the values, we'll need some more information.
(c) The maximum speed and maximum acceleration depend on the amplitude. So, once we know the amplitude, we can calculate them using good old physics formulas. You know, the ones that make us go "ugh, math."
To answer these questions, we can use Hooke's Law and the equations of motion for simple harmonic motion.
(a) To find the time it takes to reach the equilibrium position again, we need to find the period of the motion. The period is the time it takes for one complete cycle.
The period T of an object undergoing simple harmonic motion can be determined using the formula:
T = 2π√(m/k)
where m is the mass of the object and k is the spring constant.
Since the mass of the object is 1.25 kg and the spring constant (k) is not given, we cannot calculate the exact time. We need the value of k to proceed with the calculation.
(b) To write an equation giving the position y as a function of time t, we can use the equation for simple harmonic motion:
y(t) = A cos(ωt + φ)
where A is the amplitude (max displacement), ω is the angular frequency, t is the time, and φ is the phase constant.
To find A, we can use the initial displacement of the spring, which is 0.215 m. So, A = 0.215 m.
To find ω, we can use the formula:
ω = √(k/m)
Again, we need the value of k to calculate the exact angular frequency.
(c) To find the maximum speed and maximum acceleration, we can use the following relationships:
Maximum speed (v_max) occurs at the equilibrium position (y = 0), where the velocity is maximum and the acceleration is zero.
Maximum acceleration (a_max) occurs at the extreme positions, where the displacement is maximum and the acceleration is maximum.
Unfortunately, without knowing the spring constant (k), we cannot determine the maximum speed and maximum acceleration.
To solve these questions completely, we need the value of the spring constant (k).
To answer these questions, we need to use Hooke's Law and the equations for simple harmonic motion. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement or stretch of the spring.
Given information:
Mass (m) = 1.25 kg
Stretch (x1) = 0.215 m
Additional stretch (x2) = 0.130 m
(a) How long does it take to reach the equilibrium position again?
To find the time it takes for the mass to reach the equilibrium position, we need to calculate the spring constant (k) using Hooke's Law. Hooke's Law is expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement.
We can rearrange the equation to solve for k: k = -F/x
The force (F) can be calculated using the weight formula: F = mg
Plugging in the values, we have:
F = (1.25 kg)(9.8 m/s^2) = 12.25 N
k = -F/x1 = -12.25 N / 0.215 m = -56.977 N/m (rounded to three decimal places)
The angular frequency (ω) of the mass-spring system is given by ω = √(k/m)
ω² = k/m
ω² = (-56.977 N/m) / (1.25 kg)
ω² ≈ -45.5824 N/kg
ω ≈ √(-45.5824) ≈ 6.75 rad/s (rounded to two decimal places)
The time period (T) of the oscillation, which is the time taken to complete one full cycle, can be calculated using the formula T = 2π/ω.
T = 2π/6.75 ≈ 0.935 seconds (rounded to three decimal places)
Therefore, it takes approximately 0.935 seconds for the mass to reach the equilibrium position again.
(b) Write an equation giving its position y as a function of time t.
The equation for simple harmonic motion is given by: y(t) = A*cos(ωt + φ)
Where:
y(t) = position as a function of time
A = amplitude = x1 (initial stretch)
ω = angular frequency = 6.75 rad/s (from part a)
t = time
φ = phase constant
Since the mass was released from the additional stretch (x2), the initial phase constant φ can be determined using the equation: φ = -ωt
Therefore, the equation for the position is: y(t) = x1 * cos(ωt - ωt)
(c) What will be its maximum speed and maximum acceleration?
The maximum speed (Vmax) occurs at the equilibrium position, where the displacement is zero (y = 0). The maximum speed is equal to the angular frequency multiplied by the amplitude (Vmax = ω * A). Plugging in the values:
Vmax = 6.75 rad/s * 0.215 m ≈ 1.452 m/s (rounded to three decimal places)
The maximum acceleration (amax) is given by the formula amax = ω² * A. Plugging in the values:
amax = (6.75 rad/s)² * 0.215 m ≈ 9.761 m/s² (rounded to three decimal places)
Therefore, the maximum speed is approximately 1.452 m/s, and the maximum acceleration is approximately 9.761 m/s².