The function x = (6.0 m) cos [(3π rad/s)t+π/3 rad] gives the simple harmonic motion of a body. The magnitude of the acceleration at an endpoint is _____ m/s². Round to the nearest whole number.
Question
The function x = (6.0 m) cos [(3π rad/s)t+π/3 rad] gives the simple harmonic motion of a body. The magnitude of the acceleration at an endpoint is _____ m/s². Round to the nearest whole number.
Solution
The acceleration of a body in simple harmonic motion is given by the formula a = -ω²x, where ω is the angular frequency and x is the displacement.
In this case, the displacement x is given by the function x = (6.0 m) cos [(3π rad/s)t+π/3 rad]. The angular frequency ω is the coefficient of t in the argument of the cosine function, which is 3π rad/s.
At an endpoint of the motion, the displacement x is at its maximum or minimum, which is the amplitude of the motion. The amplitude is the coefficient of the cosine function, which is 6.0 m.
Substituting these values into the formula for acceleration gives a = -(3π rad/s)² * 6.0 m = -54π² m/s².
The magnitude of the acceleration is the absolute value of this, which is 54π² m/s².
To round this to the nearest whole number, we calculate the numerical value of 54π², which is approximately 536.5. Rounding to the nearest whole number gives 537 m/s².
So, the magnitude of the acceleration at an endpoint is 537 m/s².
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