Answer :
Answer:
Time taken is 0.087 s
Solution:
As per the question:
Time period, T = 0.300 s
Amplitude, A = 6.00 cm
Now,
To calculate the time taken:
For SHM, we know that:
[tex]x = Acos\omega t[/tex] (1)
At x = 6.00 cm, the object comes to rest instantaneously at times t = 0.00 s
Thus from eqn (1), for x = 6.00 cm:
[tex]6.00 = 6.00cos\omega t[/tex]
[tex]cos\omega t = 1[/tex]
[tex]\omega t = cos^{- 1}(1)[/tex]
[tex]\omega t = 0[/tex]
Thus at t = 0.00 s, x = 6.00 cm
Now,
Using eqn (1) for x = - 1.50 cm:
[tex]- 1.50 = 6.00cos\omega t'[/tex]
[tex]cos\omega t' = -0.25[/tex]
We know that:
[tex]\omega = \frac{2\pi}{T}[/tex]
Thus
[tex]\frac{2\pi}{0.300} t' = cos^{- 1}(0.25)[/tex]
[tex]t' = 0.087\ s[/tex]
Time taken by the object in moving from x = 6.00 cm to x = 1.50 cm:
t' - t = 0.087 - 0.00 = 0.087 s
Final answer:
To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we use the equation for simple harmonic motion. The time it takes is half of the period, which is 0.150 s.
Explanation:
To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we need to use the equation for simple harmonic motion (SHM). The equation is given by x = A * cos(2π/T * t + φ), where x is the position, A is the amplitude, T is the period, t is the time, and φ is the phase shift.
First, we need to determine the phase shift. At t = 0, the object is at rest at x = 6.00 cm. This means the phase shift is 0, because the cosine function is at a maximum at t = 0.
Next, we can plug in the values into the equation. The amplitude A is 6.00 cm and the period T is 0.300 s. We want to find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm. In SHM, the object goes from -A to A and back in one period, so the time it takes to go from x = 6.00 cm to x = -1.50 cm is half of the period. Therefore, the time is 0.300 s / 2 = 0.150 s.
Learn more about Time calculation in Simple Harmonic Motion here:
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