A 45 g mass is attached to a massless spring and allowed to oscillate around an equilibrium according to:
y(t) = 1.3 * sin( 0.6 * t ) where y is measured in meters and t in seconds.
(a) What is the spring constant in N/m?
k = N/m
.162 NO
HELP: You are given m, the mass. What other quantity appears in the equation involving k, the spring constant, and m?
HELP: You are given the equation of motion
y(t) = A * sin( ω * t )
Now can you find the missing quantity?
To find the missing quantity and determine the spring constant (k) in N/m, we need to identify the relationship between the equation of motion and the spring constant.
In the equation of motion for simple harmonic motion (SHM):
y(t) = A * sin(ω * t)
Where:
- y(t) represents the displacement of the mass at time t
- A represents the amplitude of the oscillation
- ω represents the angular frequency of the oscillation
By comparing this equation to the given equation, we can see that:
- The amplitude (A) in the given equation is 1.3 meters.
- The angular frequency (ω) in the given equation is 0.6.
Now, let's recall the relationship between the angular frequency (ω) and the spring constant (k) in SHM:
ω = √(k/m)
Where:
- ω is the angular frequency
- k is the spring constant
- m is the mass attached to the spring
We are given the mass (m), which is 45 grams. However, for the equation, we need to convert it to kilograms by dividing by 1000:
m = 45 g / 1000 = 0.045 kg
Now, we can rearrange the equation to solve for the spring constant (k):
k = ω^2 * m
Plugging in the values we know:
k = (0.6)^2 * 0.045
Calculating this:
k ≈ 0.0162 N/m
So, the spring constant (k) is approximately 0.0162 N/m.
Therefore, the correct answer is k = 0.0162 N/m.