A mass m = 5.7kg is attached to two springs as it slides along a frictionless floor, while the springs are fastened to two walls. The springs both have
k = 420 N/m and are both in their relaxed states (unstretched and uncompressed) when the mass is centered between the two walls. What is the frequency of this simple harmonic oscillator?
To find the frequency of the simple harmonic oscillator, we need to first determine the effective spring constant (k_eff) for the system.
The effective spring constant is given by the formula:
1/k_eff = 1/k_total = 1/k_1 + 1/k_2
Where k_1 and k_2 are the spring constants for the individual springs.
Given that k_1 = k_2 = 420 N/m, we can substitute these values into the formula:
1/k_eff = 1/420 N/m + 1/420 N/m
Simplifying the expression:
1/k_eff = 2/420 N/m
Now, we can find the value of k_eff:
k_eff = 420 N/m/2
k_eff = 210 N/m
The frequency (f) of a simple harmonic oscillator is given by the formula:
f = (1/2π) * √(k_eff / m)
Substituting the values of k_eff and m into the formula:
f = (1/2π) * √(210 N/m / 5.7 kg)
Calculating the frequency:
f ≈ (1/2π) * √(36.842/5.7)
f ≈ (1/2π) * √6.463)
f ≈ (1/2π) * 2.54
f ≈ 0.405 Hz
Therefore, the frequency of this simple harmonic oscillator is approximately 0.405 Hz.
To find the frequency of a simple harmonic oscillator, we can use the formula:
f = (1 / 2π) * √(k / m)
where
f is the frequency,
k is the spring constant, and
m is the mass.
In this case, the mass (m) is given as 5.7 kg and the spring constant (k) is given as 420 N/m.
Plugging in these values into the formula, we have:
f = (1 / 2π) * √(420 / 5.7)
Now, we can calculate the frequency.