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?
I've been using omega, but to solve for T but I'm not getting the right answer.
To solve for the spring constant, we need to use the equation of motion for a mass-spring system: y(t) = Acos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase constant.
Comparing this with the given equation, y(t) = 1.3 * sin(0.6 * t), we can see that the angular frequency is given by ω = 0.6 rad/s.
Now, recall the relationship between ω and the spring constant k: ω = sqrt(k/m), where m is the mass attached to the spring.
To find the missing quantity, we need to determine the mass. In this case, the mass is given as 45 grams, which we need to convert to kilograms by dividing by 1000. Thus, m = 45 g / 1000 = 0.045 kg.
Now we can solve for the spring constant k by rearranging the equation ω = sqrt(k/m).
ω^2 = k/m
k = ω^2 * m
Substituting the values, k = (0.6 rad/s)^2 * 0.045 kg = 0.0162 N/m.
So, the spring constant is k = 0.0162 N/m.
Please note that the value you provided (0.162 N/m) is incorrect.