A spring stretches by 0.0226 m when a 4.39-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 7.35 Hz?
7.35 = 0.0226(4.39x)
Did I set this up correctly?
The spring constant is
k =mg/Δx=4.39•9.8/0.0226 =1903 N/m.
Natural frequency is
f=sqrt(k/m) sqrt (1903/4.39) = 20.8 Hz
No, your setup is not correct. In order to calculate the mass that should be attached to the spring, you need to use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from its equilibrium position.
The equation for Hooke's Law is given by:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
To find the spring constant, you need to use the given information that the spring stretches by 0.0226 m when a 4.39-kg object is suspended from its end. The force exerted by the object can be calculated using the equation:
F = mg
where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we have:
F = (4.39 kg)(9.8 m/s²) = 43.082 N
Now, rearrange Hooke's Law equation to solve for the spring constant:
k = -F/x = -43.082 N/0.0226 m = -1901 N/m
Now that you have the spring constant, you can use the equation for the frequency of vibration of a mass-spring system:
f = 1/(2π) * sqrt(k/m)
Rearranging this equation, we can solve for the mass:
m = k/(4π²f²)
Substituting the given frequency, we have:
m = (-1901 N/m) / (4π² * (7.35 Hz)²)
Calculate this expression to find the mass that should be attached to the spring to achieve a frequency of 7.35 Hz.