VERTICAL SPRING
A spring of negligible mass, spring constant k = 89 N/m, and natural length l = 1.4 m is hanging vertically. This is shown in the left figure below where the spring is neither stretched nor compressed. In the central figure, a block of mass M = 2 kg is attached to the free end. When equilibrium is reached (the block is at rest), the length of the spring has increased by d1 with respect to l. We now lower the block by an additional d2 = 0.4 m as shown in the right figure below. At t=0 we release it (zero speed) and the block starts to oscillate. Take g=9.81 m/s2
(a) Find d1.
(b) What is the frequency (Hz) of the oscillations?
(c) What is the length of the spring when the block reaches its highest point during the oscillations?
(d) What is maximum speed of the block?
To solve this problem, we need to apply the principles of Hooke's law and simple harmonic motion.
(a) To find d1, we need to consider the equilibrium position of the spring when the block is at rest. In this position, the weight of the block is balanced by the restoring force of the spring.
The weight of the block (W) can be calculated using the formula:
W = m * g
where m is the mass of the block (2 kg) and g is the acceleration due to gravity (9.81 m/s^2). Therefore:
W = 2 kg * 9.81 m/s^2 = 19.62 N
At equilibrium, the elongation of the spring (d1) is equal to the displacement caused by the weight of the block. According to Hooke's law, the restoring force (F) of the spring is given by:
F = k * d1
where k is the spring constant (89 N/m). Equating the weight of the block and the restoring force, we have:
19.62 N = 89 N/m * d1
Solving for d1, we get:
d1 = 19.62 N / (89 N/m) = 0.22 m
Therefore, d1 = 0.22 m.
(b) The frequency (f) of the oscillations can be found using the formula:
f = 1 / T
where T is the period of the oscillations. The period is the time taken for one complete oscillation. In the case of a mass-spring system, the period is given by:
T = 2π * √(m / k)
where m is the mass of the block and k is the spring constant. Substituting the given values, we have:
T = 2π * √(2 kg / 89 N/m) ≈ 0.895 s
To find the frequency (f), we can substitute the value of T into the formula for frequency:
f = 1 / T ≈ 1.12 Hz
Therefore, the frequency of the oscillations is approximately 1.12 Hz.
(c) To find the length of the spring when the block reaches its highest point during the oscillations, we need to consider the amplitude of the oscillations. The amplitude (A) is equal to the maximum displacement from the equilibrium position.
In this case, the amplitude is equal to the sum of d1 and d2:
A = d1 + d2 = 0.22 m + 0.4 m = 0.62 m
Therefore, the length of the spring when the block reaches its highest point is the natural length of the spring (1.4 m) plus the amplitude:
Length = l + A = 1.4 m + 0.62 m = 2.02 m
The length of the spring when the block reaches its highest point during the oscillations is approximately 2.02 m.
(d) The maximum speed of the block can be determined by using energy conservation principles. At the highest point of its motion, the block momentarily stops moving before reversing its direction. This means that all of the potential energy it had at the highest point is converted into kinetic energy.
The potential energy (PE) at the highest point is given by:
PE = m * g * h
where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the highest point. In this case, h is equal to the amplitude (A) calculated in part (c).
Using the given values, we have:
PE = 2 kg * 9.81 m/s^2 * 0.62 m = 12.18 J
At the highest point, the kinetic energy (KE) is zero since the block momentarily stops moving. Therefore, the total mechanical energy (E) is equal to the potential energy at that point:
E = PE = 12.18 J
The maximum speed (v) of the block can be calculated using the formula for kinetic energy:
KE = 1/2 * m * v^2
Substituting the given values, we can solve for v:
12.18 J = 1/2 * 2 kg * v^2
v^2 = 12.18 J / (1 kg * 2)
v^2 = 6.09 J/kg
v ≈ √6.09 J/kg ≈ 2.47 m/s
Therefore, the maximum speed of the block is approximately 2.47 m/s.