A steel ball is hung from a vertical ideal spring where it oscillates in simple harmonic motion with a period T . At time t = 0 s, the ball is at its maximum displacement, A , from its equilibrium position. In terms of the period, at what time will the ball be at y = 0.75 A ? a) 0.33 T b) 0.25 T c) 0.19 T d) 0.15 T e) 0.12 T
0.12
0.15
To determine the time at which the ball will be at a displacement of 0.75A from its equilibrium position, we need to understand the relationship between the displacement and time in simple harmonic motion.
In simple harmonic motion, the displacement, velocity, and acceleration of an object can be represented by sinusoidal functions. For a vertical ideal spring, the displacement as a function of time is given by:
y(t) = A * sin(2πt / T)
Where:
y(t) is the displacement of the ball at time t,
A is the amplitude (maximum displacement) of the ball,
T is the period of the motion, and
t is the time.
Given that we want to find the time at which the displacement is 0.75A, we can set y(t) = 0.75A:
0.75A = A * sin(2πt / T)
To solve for t, we need to solve this equation for t. Let's simplify the equation:
0.75 = sin(2πt / T)
To find the value of t, we can take the inverse sine of both sides:
sin^(-1)(0.75) = 2πt / T
Since we are only interested in the time at which the ball is at y = 0.75A, we can ignore the multiple revolutions. So, we have:
sin^(-1)(0.75) = 2πt / T
To find t, we can rearrange the equation:
t = (T / 2π) * sin^(-1)(0.75)
Now, we can substitute the values given in the answer choices and see which one matches:
a) 0.33T
b) 0.25T
c) 0.19T
d) 0.15T
e) 0.12T
Using a calculator, we can calculate the value of (T / 2π) * sin^(-1)(0.75) and see which answer choice matches. Let's calculate:
t ≈ (T / 2π) * sin^(-1)(0.75) ≈ 0.19T
Therefore, the correct answer is c) 0.19T.
To find the time at which the ball will be at y = 0.75A, we need to understand the relationship between time and displacement in simple harmonic motion.
In simple harmonic motion, the displacement of the object can be described using the equation:
y = A * cos(ωt + φ)
Where:
- y is the displacement of the ball from its equilibrium position.
- A is the amplitude of the motion.
- ω is the angular frequency, given by ω = 2π/T, where T is the period of the motion.
- t is the time.
- φ is the phase constant.
From the given information, we know that the ball is at its maximum displacement, A, at t = 0s. This means that at t = 0s, y = A.
Now, we need to find the time at which y = 0.75A. Substituting this value into the equation, we have:
0.75A = A * cos(ωt + φ)
Dividing both sides by A, we get:
0.75 = cos(ωt + φ)
To find the time, we need to find the angle (ωt + φ) for which the cosine function equals 0.75.
Using a trigonometric identity, we know that cos(arccos(x)) = x. In this case, x = 0.75. So, we can rewrite our equation as:
arccos(0.75) = ωt + φ
Now, remembering that ω = 2π/T, we can substitute the expression for ω:
arccos(0.75) = (2π/T)t + φ
To find the time, we need to isolate t. Rearranging the equation, we have:
t = (T/(2π)) * (arccos(0.75) - φ)
We can simplify this formula by recognizing that at t = 0s, y = A. This corresponds to the phase constant φ being equal to 0.
So, we are left with:
t = (T/(2π)) * arccos(0.75)
Now we can calculate the value of arccos(0.75) using a calculator or online tool:
arccos(0.75) ≈ 0.7227
Now we substitute this value into the equation:
t = (T/(2π)) * 0.7227
Finally, we simplify further:
t = 0.1153T
Therefore, the ball will be at y = 0.75A at approximately 0.1153 times the period T.
Comparing this result with the provided options, we can see that the closest value is e) 0.12 T.