A 0.32 kg block oscillates back andforth along a straight line on a frictionless

Question

A 0.32 kg block oscillates back andforth along a straight line on a frictionless horizontal surface.Its displacement from the origin is given by the followingequation. x = (15cm)cos[(12 rad/s)t +/2 rad)]
What is the maximum speed acquired by the block?
cm/s
At what value of x does this occur?
cm

What is the magnitude of the maximum acceleration of the block?
cm/s2
At what values of x does this occur? (Type your answers inany order. If there is only one answer, type 'none' in the secondbox.)
cm
cm

What force, applied to the block, results in the givenoscillation?
N/cm)x x = (15cm)cos[(12 rad/s)t +/2 rad)]
What is the maximum speed acquired by the block?
cm/s
At what value of x does this occur?
cm

What is the magnitude of the maximum acceleration of the block?
cm/s2
At what values of x does this occur? (Type your answers inany order. If there is only one answer, type 'none' in the secondbox.)
cm
cm

What force, applied to the block, results in the givenoscillation?
N/cm)x

Explanation / Answer

to answer this problem you must know the mechanics of the waveequations x = (xmax)cos[t +] with this you take thederivative to get v= -xmaxsin(t +) so max speed is xmax times xmax*=15*12= 180cm/s for part b the max velocity will be when the block is at thecenter of the oscilation so if the oscilator oscilates from -15 to 15 the max speedwill be at 0 to get max acceration you take the derivative of the velocitygraph v= -xmaxsin(t +) a=-xmax2cos(t +) so max accerlation will be atxmax2 xmax2=15*122=2160cm/s2=.216m/s2 and it's max when you get to the ends so at -15 and 15 and for the last part F=ma so F=(.38)(.216m/s2)= .08208 N and it's max when you get to the ends so at -15 and 15 and for the last part F=ma so F=(.38)(.216m/s2)= .08208 N

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