An oscillator with a mass of 281 g has a speed of 119.1 cm/s when its displacement is 2.80 cm and 34.2 cm/s when its displacement is 7.80 cm. What is the oscillator's maximum speed (in m/s)?
To find the oscillator's maximum speed, we need to analyze the given data points and determine the pattern.
Let's denote the maximum speed as vmax.
From the given data, we have the following information:
- At displacement x1 = 2.80 cm, the speed v1 = 119.1 cm/s.
- At displacement x2 = 7.80 cm, the speed v2 = 34.2 cm/s.
We can use the formula for potential energy (PE) and kinetic energy (KE) to link displacement and speed:
PE = KE
For a mass-spring oscillator, the potential energy (PE) is given by:
PE = (1/2) k x^2
where k is the spring constant and x is the displacement.
The kinetic energy (KE) is given by:
KE = (1/2) m v^2
where m is the mass and v is the velocity.
Since the oscillator starts from a displacement of 0, the total mechanical energy (E) is constant:
E = PE + KE = constant
For the maximum speed, the displacement is the maximum and the potential energy is zero.
Therefore, we have:
E = (1/2) k x^2 + (1/2) m vmax^2 = 0
Solving for vmax^2, we get:
vmax^2 = -(k/m) x^2
Now, let's calculate vmax using the given data.
First, convert the mass to kilograms:
mass = 281 g = 0.281 kg
Next, convert the displacements and speeds to meters:
x1 = 2.80 cm = 0.0280 m
x2 = 7.80 cm = 0.0780 m
v1 = 119.1 cm/s = 1.191 m/s
v2 = 34.2 cm/s = 0.342 m/s
Now, we can use the equation vmax^2 = -(k/m) x^2 to solve for vmax:
vmax^2 = -(k/m) x^2
v1^2 = -(k/m) x1^2
(1.191 m/s)^2 = -(k/0.281 kg) (0.0280 m)^2
Solving this equation for k/m, we find:
k/m = -[(1.191 m/s)^2 / (0.0280 m)^2] * (0.281 kg)
k/m = -144.438
Now, we can solve for vmax^2 using the k/m value we found:
vmax^2 = -(k/m) x2^2
vmax^2 = -(-144.438) (0.0780 m)^2
vmax^2 = 0.16998
Finally, taking the square root of vmax^2, we find:
vmax = √(0.16998) = 0.4118 m/s
Therefore, the oscillator's maximum speed is approximately 0.4118 m/s.