A 0.480-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.480-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.6 cm. (Assume the position of the object is at the origin at t 0.) (a) Calculate the maximum value of its speed. cm/s (b) Calculate the maximum value of its acceleration cm/s (c) Calculate the value of its speed when the object is 8.60 cm from the equilibrium position cm/s (d) Calculate the value of its acceleration when the object is 8.60 cm from the equilibrium position. cm/s (e) Calculate the time interval required for the object to move from x 0 to x 6.60 cm.

Explanation / Answer

a)
angular freqeuncy=sqrt(spring constant/mass)=sqrt(8/0.48)=4.0825 rad/sec
let position at any time t is given by

x(t)=0.126*sin(4.0825*t)

then v(t)=dx/dt=0.5144*cos(4.0825*t)

acceleration=a(t)=dv/dt=-2.1*sin(4.0825*t)

maximum value of speed=0.5144 m/s

b)maximum value of acceleration=2.1 m/s^2

c)let at time T, position is 8.6 cm

then 0.086=0.126*sin(4.0825*T)

==>T=0.184 seconds

then speed=0.5144*cos(4.0825*0.184)=0.376 m/s

d)acceleration at T=0.184 seconds =-2.1*sin(4.0825*0.184)=-1.4333 m/s^2

e)let at time T, x(t)=6.6 cm

==>0.066=0.126*sin(4.0825*T)

==>T=0.135 seconds

hence time interval to go from x=0 to x=6.6 cm is 0.135 seconds

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