A mass of 0.40 kg,attached to a spring with a spring constant of 40 N/m,is set into simple harmonic motion.What is the magnitude of the acceleration of the mass when at its maximum displacement of 0.25 m from the equilibrium position? A)zero B)13 m/s² C)6.3 m/s² D)25 m/s² E)50 m/s²

Question

Quizplus · Accepted Answer

At maximum displacement, the mass is momentarily at rest before starting to move in the opposite direction, therefore, the acceleration is maximum.

Using the equation for simple harmonic motion:

x = A*cos(wt)

where x = displacement, A = amplitude, w = angular frequency

The maximum displacement (A) is 0.25 m.

The angular frequency (w) is given by:

w = sqrt(k/m)

where k = spring constant and m = mass

w = sqrt(40/0.40) = 10 rad/s

At the maximum displacement, the velocity is zero, so the acceleration (a) is given by:

a = -w^2*A

a = -(10)^2*(0.25)

a = -25 m/s^2

Taking the magnitude of the acceleration, we get:

|a| = 25 m/s^2

Therefore, the correct answer is D) 25 m/s^2.

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