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
.016 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?
HELP: Be careful of the units of m.
To find the spring constant (k), we can use the equation of motion for simple harmonic motion:
y(t) = A * sin(ω * t)
In this equation, A represents the amplitude of the motion, which is given as 1.3 meters. The quantity ω represents the angular frequency, which is related to the spring constant (k) and the mass (m).
Comparing the given equation with the equation of motion, we can see that ω = 0.6 rad/s.
Now we can use the relationship between ω, k, and m to find the spring constant. The equation is:
ω = sqrt(k/m)
Rearranging the equation, we get:
k = ω^2 * m
Substituting the given values, we have:
k = (0.6 rad/s)^2 * 0.045 kg
Calculating this expression, we find:
k ≈ 0.0162 N/m
Therefore, the spring constant (k) is approximately 0.0162 N/m.