a 0.25 kg mass at the end of a spring oscillates 2.2 times per second with an amplitude of 0.15m. Determine the speed when it passes the equilibrium point and the speeed when it is 0.1 m from equilibrium
First we need to find the angular frequency (ω) using the given formula:
ω = 2π * frequency
where frequency is given to be 2.2 oscillations per second.
ω = 2π * 2.2 ≈ 13.82 rad/s
Now we can determine the spring constant (k) using the formula:
k = m * ω²
where m is the mass (0.25 kg) and ω is the angular frequency (13.82 rad/s).
k = 0.25 * (13.82)² ≈ 47.60 N/m
Next, we can find the maximum speed (v_max) using the formula:
v_max = A * ω
where A is the amplitude (0.15 m) and ω is the angular frequency (13.82 rad/s).
v_max = 0.15 * 13.82 ≈ 2.07 m/s
When the mass passes through the equilibrium point, the speed is the maximum speed, v_max.
Speed at equilibrium point = 2.07 m/s
Now, let's determine the speed when the mass is 0.1 m from the equilibrium. Using the formula:
v² = ω² * (A² - x²)
where v is the speed, A is the amplitude (0.15 m), ω is the angular frequency (13.82 rad/s), and x is the displacement from equilibrium (0.1 m).
v² = (13.82)² * (0.15² - 0.1²)
v² = 190.99 * (0.0225 - 0.01)
v² = 190.99 * 0.0125
v² ≈ 2.39
Taking the square root:
v ≈ √2.39 ≈ 1.54 m/s
Speed at 0.1 m from equilibrium = 1.54 m/s
To determine the speed when the mass passes the equilibrium point, we need to know its maximum velocity. The maximum velocity occurs when the mass is at the amplitude of oscillation.
The formula for the maximum velocity of a mass-spring system is given by:
V_max = A * ω
where V_max is the maximum velocity, A is the amplitude of oscillation, and ω is the angular frequency.
To calculate the angular frequency, we can use the formula:
ω = 2πf
where f is the frequency of oscillation.
Given that the mass oscillates 2.2 times per second, we have:
f = 2.2 Hz
Using the formula for angular frequency:
ω = 2π * 2.2 Hz = 13.84 rad/s
Now we can calculate the maximum velocity:
V_max = 0.15m * 13.84 rad/s ≈ 2.076 m/s
Therefore, the speed when the mass passes the equilibrium point is approximately 2.076 m/s.
To determine the speed when the mass is 0.1 m from equilibrium, we can use the formula for velocity at a specific displacement in simple harmonic motion:
V = ω * √(A^2 - x^2)
where V is the velocity, x is the displacement from equilibrium, and all other variables are as previously defined.
Given that the displacement is 0.1 m, we have:
x = 0.1 m
Using the previous value for angular frequency:
V = 13.84 rad/s * √(0.15^2 - 0.1^2) ≈ 7.598 m/s
Therefore, the speed when the mass is 0.1 m from equilibrium is approximately 7.598 m/s.