An object executes simple harmonic motion with an amplitude A.
a) At what values of its position does its speed equal half its maximum
To determine at what values of its position the speed of an object executing simple harmonic motion equals half its maximum, we need to understand the relationship between the position, velocity, and acceleration in simple harmonic motion.
In simple harmonic motion, the relationship between position, velocity, and acceleration can be represented by the equation:
x(t) = A * cos(ωt + φ),
where:
- x(t) represents the position of the object at time t,
- A is the amplitude of the motion,
- ω is the angular frequency of the motion, given by ω = 2πf, where f is the frequency,
- φ is the phase constant.
The velocity and acceleration of the object can be obtained by differentiating the position equation:
v(t) = dx(t)/dt = -A * ω * sin(ωt + φ),
a(t) = dv(t)/dt = -A * ω^2 * cos(ωt + φ),
where:
- v(t) represents the velocity of the object at time t,
- a(t) represents the acceleration of the object at time t.
Now, let's find the values of position (x) where the speed (|v|) is half its maximum value.
The speed of the object is given by the absolute value of its velocity, |v(t)| = |-A * ω * sin(ωt + φ)| = A * ω * |sin(ωt + φ)|.
Half of the maximum speed is given by (1/2) * A * ω.
To find the values of position where the speed is half its maximum, we need to solve the equation:
A * ω * |sin(ωt + φ)| = (1/2) * A * ω.
We can simplify this equation to:
|sin(ωt + φ)| = 1/2.
To find the values of position for which the above equation is satisfied, we need to determine the points where the absolute value of the sine function is equal to 1/2.
The absolute value of the sine function is equal to 1/2 at two specific values:
1) sin(ωt + φ) = 1/2,
2) sin(ωt + φ) = -1/2.
We can solve these two equations to find the corresponding values of position.
To determine the values of t that satisfy the first equation, sin(ωt + φ) = 1/2, we need to solve for t using the inverse sine function:
ωt + φ = arcsin(1/2) = π/6 or 5π/6.
Similarly, for the second equation, sin(ωt + φ) = -1/2, we have:
ωt + φ = arcsin(-1/2) = -π/6 or -5π/6.
Now, we can solve for the position values (x) associated with these values of t using the position equation:
x(t) = A * cos(ωt + φ).
Plug in the values of t to find the corresponding positions x:
1) At t = (π/6 - φ) / ω: x = A * cos(π/6),
2) At t = (5π/6 - φ) / ω: x = A * cos(5π/6),
3) At t = (-π/6 - φ) / ω: x = A * cos(-π/6),
4) At t = (-5π/6 - φ) / ω: x = A * cos(-5π/6).
Therefore, the positions where the speed of the object is half its maximum occur at x = A * cos(π/6), x = A * cos(5π/6), x = A * cos(-π/6), and x = A * cos(-5π/6).