A 166 g mass is attached to a horizontally mounted spring. It is pulled out to stretch the spring and then released. You determine the period of the oscilliation to be 1.0 seconds. What is the acceleration of the mass (magnitude) when the spring is stretched 33 cm during the oscillation?
To find the acceleration of the mass when the spring is stretched 33 cm during the oscillation, we can use Hooke's Law and the formula for the period of oscillation.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from equilibrium. Mathematically, it can be expressed as:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from equilibrium.
In this case, the mass is attached to a horizontally mounted spring and is oscillating, so we can write:
ma = -kx
where m is the mass of the object and a is acceleration.
The period of oscillation is given by the formula:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
We can rearrange the equation for period to find the spring constant:
k = (4π²m) / T²
Now that we have the spring constant, we can substitute it into the equation for the force exerted by the spring:
F = -kx
Finally, we can find the acceleration by using Newton's second law:
a = F / m
Substituting everything in, we have:
a = -kx / m
Now we can plug in the values:
m = 166 g = 0.166 kg (convert grams to kilograms)
x = 33 cm = 0.33 m (convert centimeters to meters)
T = 1.0 s
First, calculate the spring constant:
k = (4π² * 0.166) / (1.0)²
Then, calculate the acceleration:
a = (-k * 0.33) / 0.166
Solving for a gives you the magnitude of the acceleration of the mass when the spring is stretched 33 cm during the oscillation.