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
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
1 answer