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
The time it takes the object to go from x = 6.00 cm to x = -1.50 cm is 0.063 s.
The given parameters;
The general wave equation is given as;
[tex]y = A cos\ \omega t\\\\6 = 6 cos\ \omega (t)\\\\cos \ \omega t = \frac{6}{6} \\\\cos \ \omega t = 1\\\\\omega t = cos^{-1} \\\\\omega t = 0[/tex]
Thus, at x = 6 cm, t = 0 s
When x = -1.5 cm, the time t = ?
[tex]1.5 \ = 6cos\ \omega t_2\\\\cos\ \omega t_2 = \frac{1.5}{6} \\\\cos\ \omega t_2 = 0.25\\\\\omega t_2 = cos^{-1} (0.25)\\\\\omega t_2 = 1.32\\\\t_2 = \frac{1.32}{\omega} \\\\t_2 = \frac{1.32 \ \times T}{2\pi} \\\\t_2 = \frac{1.32 \times 0.3}{2\pi } \\\\t_2 = 0.063 \ s[/tex]
The time taken from 6 cm to 1.5 cm
t = 0.063 s - 0
t = 0.063 s.
Thus, the time it takes the object to go from x = 6.00 cm to x = -1.50 cm is 0.063 s.
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