A block with mass 1.973 kg sliding on a frictionless surface is attached to a horizontal spring with spring constant 24.33 N/m at the equilibrium position of the spring. The block is then pulled a distance 16.13 cm from the equilibrium point and released from rest. What is the distance of the block from the equilibrium point after 2.907×10-1 s?
The motion of a block on a spring can be described by simple harmonic motion. The displacement of the block from the equilibrium point at time t is given by:
x(t) = A * cos(ωt + φ)
Where x(t) is the displacement from the equilibrium point at time t, A is the amplitude of the motion, ω is the angular frequency, and φ is the phase angle. When the block is released from rest at its maximum displacement, its initial phase angle (φ) will be zero. The amplitude (A) of the motion is equal to the initial displacement, which is 0.1613 meters.
The angular frequency (ω) can be found using the equation:
ω = sqrt(k/m)
where k is the spring constant, and m is the mass of the block. Plugging in the values given in the problem:
ω = sqrt(24.33 N/m / 1.973 kg) = 3.511 rad/s
Now we can find the displacement (x) at the given time t = 2.907 * 10^-1 s. Plugging in the values into the displacement equation:
x(t) = A * cos(ωt)
x(2.907 * 10^-1 s) = 0.1613 m * cos(3.511 rad/s * 2.907 * 10^-1 s) = 0.1395 m
The distance of the block from the equilibrium point after 2.907×10^-1 s is approximately 0.1395 m or 13.95 cm.