A body of unknown mass is attached to an ideal spring with force constant 120N/m in the arrangement shown in Figure
-SP2. It is found to vibrate with a frequency of 6.00Hz. Find a) the period b) the angular frequency and c) mass of the
body.
To find the period, angular frequency, and mass of the body, we can use the formula for the frequency of oscillation of a mass-spring system:
f = 1 / T
where f is the frequency and T is the period. Rearranging this equation, we can solve for T:
T = 1 / f
a) To find the period, we can substitute the given frequency into the equation:
T = 1 / 6.00 Hz
T ≈ 0.1667 s
So, the period of the vibration is approximately 0.1667 seconds.
b) The angular frequency (ω) of the oscillation can be calculated using the formula:
ω = 2πf
where f is the frequency. Substituting the given frequency, we have:
ω = 2π * 6.00 Hz
ω ≈ 37.7 rad/s
So, the angular frequency of the vibration is approximately 37.7 radians per second.
c) To find the mass of the body, we can use the formula for the natural angular frequency (ω₀) of a mass-spring system:
ω₀ = √(k / m)
where k is the force constant of the spring and m is the mass. Rearranging this equation, we can solve for m:
m = k / ω₀²
Substituting the given force constant (k = 120 N/m) and the calculated angular frequency (ω₀ = 37.7 rad/s), we have:
m = 120 N/m / (37.7 rad/s)²
m ≈ 0.087 kg
So, the mass of the body is approximately 0.087 kilograms.