A block of mass 4kg hangs from a spring of force constant 400N/m.the block is pulled down 15cm below the equilibrium position and released. Fimd: a) amplitude, frequency and period of vibration. b) kinetic energy when the block is 10 cm above the equilibrium position. c) time it takes the block to go 12 cm below the equilibrium pOsition on the way up to 9 cm above the equilibrium position
To solve this problem, we'll use some basic principles of simple harmonic motion and energy calculations. Let's break it down step by step:
a) First, let's find the amplitude, frequency, and period of vibration:
- Amplitude (A): The amplitude is the maximum displacement from the equilibrium position.
In this case, the block is pulled down 15 cm below the equilibrium position. Since the displacement is always measured from the equilibrium position, the amplitude is 15 cm.
A = 15 cm
- Frequency (f): Frequency is the number of complete oscillations per unit of time.
We can use the equation f = 1 / T, where T is the period of vibration.
Given that the force constant, k, of the spring is 400 N/m, we can use the equation T = 2π√(m / k), where m is the mass of the block.
M = 4 kg
T = 2π√(4 kg / 400 N/m)
T = 2π√(0.01) ≈ 2π * 0.1 ≈ 0.628 s
f = 1 / T = 1 / 0.628 ≈ 1.59 Hz
The frequency is approximately 1.59 Hz.
- Period (T): The period is the time taken to complete one cycle of oscillation.
We already calculated the period using the formula T = 2π√(m / k) as approximately 0.628 s.
Therefore, the amplitude is 15 cm, the frequency is approximately 1.59 Hz, and the period is approximately 0.628 s.
b) To find the kinetic energy when the block is 10 cm above the equilibrium position, we need to calculate the velocity at that point and use the equation KE = 0.5 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
To find the velocity, we'll use the conservation of mechanical energy. At the maximum displacement, all potential energy has been converted to kinetic energy.
Potential energy at the maximum displacement (10 cm above equilibrium):
Using the equation PE = 0.5 * k * x^2, where PE is the potential energy and x is the displacement from the equilibrium position.
PE = 0.5 * 400 N/m * (0.1 m)^2 = 2 J
Since all potential energy converts to kinetic energy at the maximum displacement:
KE = PE = 2 J
So, the kinetic energy when the block is 10 cm above the equilibrium position is 2 J.
c) To find the time it takes for the block to go 12 cm below the equilibrium position on the way up to 9 cm above the equilibrium position, we need to calculate the time taken for each part separately.
Time taken to go 12 cm below the equilibrium position:
Using the equation T = 2π√(m / k), we can calculate the period.
T = 2π√(4 kg / 400 N/m) = 2π√(0.01) ≈ 2π * 0.1 ≈ 0.628 s
Time taken to go 9 cm above equilibrium position:
As it is the same distance from the equilibrium position as the amplitude, the time taken will also be the same.
Therefore, the time taken is approximately 0.628 s.
Thus, the time it takes for the block to go 12 cm below the equilibrium position on the way up to 9 cm above the equilibrium position is approximately 0.628 s.
I hope this helps! Let me know if you have any further questions.
a) are not these standard formulas?
b) You know total energy from the amplitude, and at 10 cm, you know from the force equation the PE, so then you know Ke
c) use the displacement equation to figure time for each position, the difference in time is the time taken.