A spring with a spring constant of 1.50 102 N/m is attached to a 1.2 kg mass and then set in motion.
(a) What is the period of the mass-spring system?
s
(b) What is the frequency of the vibration?
Hz
look at the units:
spring constant= N/m=(kg*m/s)/m = kg/s
to get seconds (period): I want s in the numerator. kg / (kg/s) = s
to get frequency (s^-1): I want s in the denominator: 1/period
opps!
spring constant= N/m=(kg*m/s^2)/m = kg/s^2
econds (period): I want s in the numerator. kg / (kg/s) = s^2 (square root it?)
period = T = 2Pi √(m/k)
I looked it up. my derivation with units was correct except for the fact that it's multiplied by 2 pi
To determine the period and frequency of the mass-spring system, we can use the equation:
Period (T) = 2π√(m/k)
Where:
- T is the period (in seconds)
- π is a mathematical constant (~3.14159)
- m is the mass (in kilograms)
- k is the spring constant (in N/m)
(a) To find the period:
First, we need to substitute the given values into the equation:
T = 2π√(m/k)
T = 2π√(1.2 kg / 1.50 × 102 N/m)
Now, we can calculate the period:
T ≈ 2π√(1.2 / 150)
T ≈ 2π√(0.008)
Using a calculator, we can find:
T ≈ 2π × 0.089
Finally, we can solve this to get the period:
T ≈ 0.56 s
Therefore, the period of the mass-spring system is approximately 0.56 seconds.
(b) To find the frequency:
Frequency (f) is the reciprocal of the period:
f = 1 / T
Substituting the determined period, we can calculate the frequency:
f ≈ 1 / 0.56
Simplifying this, we find:
f ≈ 1.79 Hz
Therefore, the frequency of the vibration is approximately 1.79 Hz.