A 2.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin($\omega$ t).What is the maximum acceleration of the simple harmonic motion? A) 17.5 cm/s² B) 21.2 cm/s² C) 25.2 cm/s² D) 30.5 cm/s² E) 36.7 cm/s²

Question

Quizplus · Accepted Answer

The maximum acceleration of a mass on a spring is given by the formula:

a_max = w^2 * A

where w is the angular frequency (2*pi*f) and A is the amplitude of the motion.

In this case, we can find the angular frequency by noting that the velocity function is given by:

v(t) = A*w*cos(w*t)

Comparing this to the given velocity function, we see that:

A*w = 10.0 cm/s

and

w = 2*pi*f

So,

f = w/(2*pi) = (10.0 cm/s)/(2*pi*(0.01 m)) = 1.59 Hz

Now we can find the amplitude of the motion by noting that the maximum speed is equal to the amplitude:

v_max = A*w

A = v_max/w = (10.0 cm/s)/((2*pi)*(1.59 Hz)) = 0.994 cm

Finally, we can find the maximum acceleration:

a_max = w^2*A = ((2*pi*f)^2)*(0.994 cm) = 21.2 cm/s^2

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