A 200 g block is attached to a spring. The block is pushed in 10 cm and released. An observer
with a stopwatch finds that 10 oscillations take 12.0 s.
(a) What is the spring constant?
(b) Draw an appropriately labelled position-time graph for this motion showing 2 cycles.
(c) What is the KE, PE and total energy of the block when it is 4 cm to the right of
equilibrium?
(a) To find the spring constant, we can use the formula for the period of oscillation:
T = 2π√(m/k)
Where T is the period, m is the mass of the block, and k is the spring constant.
In this case, T = 12.0 s and m = 200 g = 0.2 kg.
Rearranging the formula, we have:
k = (4π²m) / T²
Plugging in the values, we get:
k = (4π² * 0.2) / (12.0)²
k = 1.047 N/m
Therefore, the spring constant is approximately 1.047 N/m.
(b) The position-time graph for the motion can be represented as a sinusoidal wave. The x-axis represents time and the y-axis represents position.
The graph will start at the equilibrium position (x = 0) and the blocks moves to the left (negative x direction), reaches the maximum negative displacement and then starts oscillating symmetrically around the equilibrium position with decreasing amplitude.
(c) When the block is 4 cm to the right of equilibrium, it has potential energy (PE) and kinetic energy (KE).
PE = (1/2)kx²
Where k is the spring constant and x is the displacement from the equilibrium position.
In this case, x = 4 cm = 0.04 m and k = 1.047 N/m.
Plugging in the values, we have:
PE = (1/2) * 1.047 * (0.04)²
PE = 0.000334 Nm (or J)
The total energy (TE) of the block is the sum of its potential and kinetic energy:
TE = PE + KE
Since the block is at its maximum displacement, its kinetic energy is zero.
Therefore, TE = PE = 0.000334 Nm (or J)