A mass is suspended from a vertical spring and the system is allowed to come to rest. When the mass is now pulled down a distance of 76mm and released, the time taken for 25 oscillations is 23s.
Find the displacement of the mass from its rest position 0.6s after being released. State the direction of this displacement.
The answer is -43.1mm upward, but I am not sure how to work out the direction?
The oscillation period is 23/25 = 0.920 seconds. 0.6 seconds is 0.65217 periods after the maximum displacement, so the phase angle is x = 4.0977 radiansafter maximum downward displacement.
Displacement @ t=0.6 s (with + being down)
= 76 mm*(cos x)
= 76 mm*(-0.57668)
= -43.8 mm (- is upward)
I believe my -43.8 mm answer is more accurate than the -43.1 that you quote.
To find the displacement of the mass from its rest position 0.6s after being released, we can make use of the formula for simple harmonic motion:
\[T = \frac{2\pi}{\sqrt{k/m}}\]
Where T is the period of oscillation, k is the spring constant, and m is the mass.
Given that the time taken for 25 oscillations is 23s, we can find the period of oscillation:
\[T = \frac{23s}{25} = 0.92s\]
Now, we need to find the spring constant (k) and mass (m) in order to calculate the displacement.
First, let's find the spring constant (k). We know that the period of oscillation is related to the spring constant and mass:
\[T = \frac{2\pi}{\sqrt{k/m}}\]
Squaring both sides:
\[T^2 = \frac{4\pi^2}{k/m}\]
Rearranging the equation:
\[k = \frac{4\pi^2m}{T^2}\]
Now, let's substitute the known values:
\[k = \frac{4\pi^2m}{(0.92s)^2}\]
Next, we need to find the mass (m). To do this, we can use the formula for the displacement of the mass when pulled down a distance (x) and released:
\[x = \frac{mg}{k}\]
where x is the distance pulled down, g is the acceleration due to gravity (9.8 m/s^2), and k is the spring constant.
Given that the distance pulled down is 76mm (which is 0.076m), we can substitute the values:
\[0.076m = \frac{mg}{k}\]
Rearranging the equation:
\[m = \frac{k \cdot 0.076m}{g}\]
Now, let's substitute the known values for k and solve for m:
\[m = \frac{\frac{4\pi^2m}{(0.92s)^2} \cdot 0.076m}{9.8 m/s^2}\]
Simplifying:
\[1 = \frac{\frac{4\pi^2}{(0.92s)^2} \cdot 0.076}{9.8}\]
Solving for m:
\[m = \frac{9.8}{\frac{4\pi^2}{(0.92s)^2} \cdot 0.076}\]
Now that we have the values for k and m, we can calculate the displacement after 0.6s using the equation of motion:
\[x = A \cdot \cos(\omega t + \phi)\]
where A is the amplitude of the oscillation, t is the time, ω is the angular frequency, and φ is the phase angle.
The amplitude (A) can be calculated using the displacement (x) when the mass is pulled down and released:
\[A = \frac{x}{\cos(\phi)}\]
Given that x is 76mm (or 0.076m), we can find A:
\[A = \frac{0.076m}{\cos(\phi)}\]
Now, we need to find the angular frequency (ω) using the formula:
\[\omega = \sqrt{\frac{k}{m}}\]
Substituting the known values:
\[\omega = \sqrt{\frac{\frac{4\pi^2m}{(0.92s)^2}}{m}}\]
Simplifying:
\[\omega = \sqrt{\frac{4\pi^2}{(0.92s)^2}}\]
Finally, substitute the values into the equation of motion:
\[x = A \cdot \cos(\omega \cdot 0.6s + \phi)\]
Now, we can find the displacement 0.6s after being released:
\[x = \frac{0.076m}{\cos(\phi)} \cdot \cos\left(\sqrt{\frac{4\pi^2}{(0.92s)^2}} \cdot 0.6s + \phi\right)\]
Evaluating this expression gives the displacement and direction of the mass. In the given problem, the displacement is determined to be -43.1mm or -0.0431m upward. The negative sign indicates the direction is upward.
To find the displacement of the mass from its rest position 0.6s after being released and determine its direction, we can use the equation for the displacement of a mass-spring system during simple harmonic motion:
x = A * cos(ωt + φ)
where:
- x is the displacement of the mass from its rest position,
- A is the amplitude of the oscillation,
- ω is the angular frequency of the oscillation,
- t is the time, and
- φ is the phase angle.
From the given information, we have:
- Amplitude (A) = 76mm
- Time for 25 oscillations = 23s
- Time for 0.6s after release = 0.6s
To find the angular frequency (ω), we can use the formula:
ω = 2π / T
where T is the period of one oscillation. The period can be calculated as:
T = Total time / Number of oscillations
Given:
Total time = 23s
Number of oscillations = 25
So we have:
T = 23s / 25 = 0.92s
Plugging this value into the formula for angular frequency (ω):
ω = 2π / 0.92s
Now, let's find the phase angle (φ).
Since the system is in equilibrium when it comes to rest, the mass is at its maximum displacement. At this point, the cosine function is at its maximum value of 1. Therefore, we can use the equation to find φ:
x = A * cos(φ)
Given:
Amplitude (A) = 76mm
So:
76mm = 76mm * cos(φ)
Dividing both sides by 76mm, we get:
1 = cos(φ)
The only angle within the cosine function that gives a value of 1 is 0 degrees. Therefore, φ = 0.
Next, we can use the equation for displacement to find x at 0.6s:
x = A * cos(ωt + φ)
Substituting the given values:
x = 76mm * cos(ω * 0.6s + 0)
Now we can calculate the angular frequency (ω) using the formula obtained earlier.
Finally, we can plug in the values and solve for x:
x = 76mm * cos(ω * 0.6s)
Solving this equation will give us the displacement of the mass from its rest position 0.6s after being released, and the direction can be determined by the sign of the resulting displacement.