In experiment is carried out to measure the spring constant of a spring. a mass of 500 g is suspended on the string. its is pulled down a small distance and the time for 20 oscillation is measured to be 34s.
a)explain why the mass performs simple harmonic motion?
b) what is the spring constant?
b/c in simple harmonic motion there is constant oscillation
What Is The Spring Constant ?
7887.68N/m
A)because an experiment is carried out to measure the spring constant of a spring.
B)k=m*number of oscillation /measured time
K=0.5*20/34=0.3N/m
c)x=mg/k=0.5*10/0.3=17m
D)i)change in gravity doesnt affect the time period of SHM.
ii)x=m*g of moon/k=0.5*1.6/0.3=2.72m
a) The mass performs simple harmonic motion because it is subject to a restoring force that is directly proportional to its displacement from the equilibrium position. In this case, the restoring force is provided by the spring, which follows Hooke's Law. When the mass is displaced from its equilibrium position and released, the spring exerts a force that restores it back towards the equilibrium position. Due to this restoring force, the mass oscillates back and forth around the equilibrium position.
b) To find the spring constant, we can use the formula for the period (T) of a mass-spring system:
T = 2π√(m/k)
where T is the period, m is the mass attached to the spring, and k is the spring constant.
Given that the period is 34 s and the mass (m) is 500 g (which can be converted to 0.5 kg), we can rearrange the formula to solve for the spring constant (k):
k = (4π²m) / T²
Substituting the values into the formula:
k = (4π² * 0.5) / (34²)
Simplifying the equation:
k ≈ 3.067 N/m
Therefore, the spring constant is approximately 3.067 N/m.