Respuesta :
Answer:
a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s
b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s
Explanation:
a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)
It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.
Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).
Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s
b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.
The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,
0.5T = 2.76 s
T = 2 × 2.76 = 5.52 s
Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as
y = A cos wt (provided that there's no phase difference, that is, Φ = 0)
A = amplitude = 0.18 m
w = (2π/5.52) = 1.138 rad/s
When y = 0.09 m, the time = t₁₂ = ?
0.09 = 0.18 Cos 1.138t₁ (angles in radians)
Cos 1.138t₁ = 0.5
1.138t₁ = arccos (0.5) = (π/3)
t₁ = π/(3×1.138) = 0.92 s
When y = -0.09 m, the time = t₂ = ?
-0.09 = 0.18 Cos 1.138t₂ (angles in radians)
Cos 1.138t₂ = -0.5
1.138t₂ = arccos (-0.5) = (2π/3)
t₂ = 2π/(3×1.138) = 1.84 s
Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s
Hope this Helps!!!
The time required for the block to travel from x = 0.180m to x = -0.180 m is 2.76 s. The time required for the block to travel from x = 0.090m to x = -0.090 m is 0.92 s
From the given information, the period of time for the system can be computed as:
[tex]\mathbf{T = \dfrac{2 \pi}{\omega}}[/tex]
T = 2(2.76)s
T = 5.52 s
Thus, the period to travel from the positive 0.180 m to -0.180 m is 2.76s since the period is not dependant on the mass of the block and the spring constant.
(b)
Taking the position with time according to the simple harmonic equation, we have:
[tex]\mathbf{x_1 = A cos (\omega t)} \\ \\ \\ \mathbf{x_1 = Acos (\dfrac{2 \pi t}{T})} \\ \\ \\ \mathbf{\dfrac{2 \pi t_1}{T}= \dfrac{\pi}{3}} \\ \\ \\ \mathbf{\dfrac{2 \pi t_2}{T}= 2\dfrac{\pi}{3}}[/tex]
Now;
[tex]\mathbf{t_1 = \dfrac{T}{6}}[/tex]
[tex]\mathbf{t_2= \dfrac{T}{3}}[/tex]
Equating both t₁ and t₂ together, we have:
[tex]\mathbf{t_2 - t_1 = \dfrac{T}{6}}}[/tex]
[tex]\mathbf{t_2 - t_1 = \dfrac{5.52 \ s}{6}}}[/tex]
[tex]\mathbf{t_2 - t_1 =0.92 \ s }}[/tex]
Therefore, we can conclude that the time required for the block to travel from x = 0.090m to x = -0.090 m is 0.92 s
Learn more about simple harmonic motion here:
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