A simple harmonic oscillator consists of a 1.1 kg block attached to a spring (k = 180 N/m). The block slides on a horizontal frictionless surface about the equilibrium point x = 0 with a total mechanical energy of 3.0 J. (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete in 15 s? (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at x = 0.14 m?
To find the answers to the questions, we can use the equations governing the motion of a simple harmonic oscillator:
(a) The amplitude can be found using the equation for total mechanical energy:
E = (1/2) k A^2
where E is the total mechanical energy and A is the amplitude.
Plugging in the given values: 3.0 J = (1/2) * 180 N/m * A^2
Solving for A: A = √( (2 * E) / k ) = √( (2 * 3.0 J) / 180 N/m ) ≈ 0.247 m
So, the amplitude of the oscillation is approximately 0.247 m.
(b) The period of the oscillation can be found using the equation:
T = 2π√(m/k)
where T is the period of oscillation, m is the mass of the block, and k is the spring constant.
Plugging in the given values: T = 2π√(1.1 kg / 180 N/m) ≈ 0.753 s
Now, we can find the number of oscillations completed in 15 s:
Number of oscillations = (total time) / (time for one oscillation)
Number of oscillations = 15 s / 0.753 s ≈ 19.89
So, the block completes approximately 19.89 oscillations in 15 s.
(c) The maximum kinetic energy attained by the block is equal to the total mechanical energy since there is no potential energy at that point.
Maximum kinetic energy = 3.0 J
So, the maximum kinetic energy attained by the block is 3.0 J.
(d) To find the speed of the block at x = 0.14 m, we can use the energy conservation equation:
E = (1/2) m v^2 + (1/2) k x^2
where v is the velocity of the block and x is its displacement from equilibrium.
Plugging in the given values: 3.0 J = (1/2) * 1.1 kg * v^2 + (1/2) * 180 N/m * (0.14 m)^2
Solving for v: v = √((2 * E - k * x^2) / m ) = √((2 * 3.0 J - 180 N/m * (0.14 m)^2) / 1.1 kg ) ≈ 0.589 m/s
So, the speed of the block at x = 0.14 m is approximately 0.589 m/s.
To solve these questions, we can use the principles of conservation of mechanical energy and the equation of motion for a simple harmonic oscillator.
(a) Amplitude of the oscillation:
The mechanical energy of the system is given by the sum of potential energy and kinetic energy. For a mass-spring system, the mechanical energy is given by:
E = (1/2)kA²
where E is the mechanical energy, k is the spring constant, and A is the amplitude of the oscillation.
Given E = 3.0 J and k = 180 N/m, we can solve for A:
3.0 = (1/2)(180)(A²)
6.0 = 180A²
A² = 6.0/180
A² = 0.0333
A ≈ √0.0333
A ≈ 0.182 m
Therefore, the approximate amplitude of the oscillation is 0.182 m.
(b) Number of oscillations in 15 seconds:
The time period of a simple harmonic oscillator can be calculated using the formula:
T = 2π√(m/k)
where T is the time period, m is the mass, and k is the spring constant.
Given m = 1.1 kg and k = 180 N/m, we can calculate the time period of the oscillator:
T = 2π√(1.1/180)
T ≈ 0.521 s
The number of oscillations in 15 seconds is:
Number of oscillations = time / T
Number of oscillations = 15 / 0.521
Number of oscillations ≈ 28.8
Therefore, the block completes approximately 28.8 oscillations in 15 seconds.
(c) Maximum kinetic energy attained by the block:
At the maximum displacement from equilibrium, all the mechanical energy of the system is converted into kinetic energy.
Therefore, the maximum kinetic energy is equal to the mechanical energy, which is 3.0 J.
Therefore, the maximum kinetic energy attained by the block is 3.0 J.
(d) Speed of the block at x = 0.14 m:
The equation of motion for a simple harmonic oscillator is:
x = A sin(ωt)
where x is the displacement, A is the amplitude, and ω is the angular frequency.
The angular frequency can be calculated using the formula:
ω = √(k/m)
Given k = 180 N/m and m = 1.1 kg, we can calculate ω:
ω = √(180/1.1)
ω ≈ 12.79 rad/s
At x = 0.14 m, we can substitute the values in the equation of motion:
0.14 = 0.182 sin(12.79t)
To find the speed, we need to differentiate the equation of motion with respect to time:
v = dx/dt = Aω cos(ωt)
At x = 0.14 m, the speed can be calculated:
v = 0.182 * 12.79 cos(12.79t)
Substituting the value of t when x = 0.14 m, we can find the speed.
Note: The value of t will depend on the phase or initial conditions, which are not given in the question.