A spring of negligible mass, spring constant k = 82 N/m, and natural length l = 1.2 m is hanging vertically. This is shown in the left side where the spring is neither stretched nor compressed. In the central one, a block of mass M = 5 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 = 0.2 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
What is the frequency (Hz) of the oscillations?
What is the length of the spring when the block reaches its highest point during the oscillations?
What is maximum speed of the block?
Please answer asap
f+0.63
A spring of negligible mass, spring constant k = 81 N/m, and natural length l = 1.5 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 = 1 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
(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 find the frequency of the oscillations, we can use the formula:
f = 1 / T
where f is the frequency and T is the period of the oscillation.
The period of a mass-spring system is given by:
T = 2π√(m/k)
where m is the mass attached to the spring and k is the spring constant. In this case, the mass is M = 5 kg and the spring constant is k = 82 N/m.
So, plugging in the values:
T = 2π√(5 / 82) = 2π√(0.06098) ≈ 0.7689 seconds
Now we can calculate the frequency:
f = 1 / T = 1 / 0.7689 ≈ 1.3009 Hz
Therefore, the frequency of the oscillations is approximately 1.3009 Hz.
To find the length of the spring when the block reaches its highest point during the oscillations, we need to consider the equilibrium position as the reference point. When the block is at its highest point, the spring is at its maximum extension, which is the sum of the equilibrium length (l) and the displacement d1 + d2.
So, the length of the spring at the highest point can be calculated as:
Length at highest point = l + d1 + d2 = 1.2 + d1 + 0.2 = 1.4 + d1 meters
To find the maximum speed of the block, we can use the conservation of mechanical energy. At the highest point of the oscillation, all the potential energy is converted to kinetic energy, so we can equate them:
Potential Energy = Kinetic Energy
At the highest point, the potential energy is given by:
Potential Energy = m * g * height
where m is the mass of the block (5 kg) and g is the acceleration due to gravity (9.81 m/s^2). The height at the highest point is the displacement d1.
So, the potential energy is:
Potential Energy = 5 * 9.81 * d1 = 49.05 * d1 J
The kinetic energy at the highest point is given by:
Kinetic Energy = (1/2) * m * v^2
where v is the maximum speed of the block. Since the block is momentarily at rest at the highest point and then reverses direction, the maximum speed is when the block is passing through the equilibrium position.
So, the kinetic energy at the highest point is:
Kinetic Energy = (1/2) * 5 * v_eq^2 = 2.5 * v_eq^2 J
Equating the potential and kinetic energies:
49.05 * d1 = 2.5 * v_eq^2
Simplifying:
v_eq^2 = 49.05 * d1 / 2.5
Taking the square root of both sides:
v_eq = √(49.05 * d1 / 2.5)
Therefore, the maximum speed of the block is approximately √(49.05 * d1 / 2.5) m/s.