A 200g mass attached to a horizontal spring oscillates at a

A 200g mass .attached to a horizontal spring oscillates at a frequency of 2.0 Hz and an amplitude of 5.0 cm. The mass is pulled 2.S cm to the right and given an initial velocity of 54.41 cm/s to the right. (Switch your calculator to radian mode If you use any Cot or Sin functions for (his part) Determine the following

Solution

mas m= 200g

frequency f= 2Hz

amplitude A= 5.0 cm

equation of motion is given by

x= ASin(2f t) i.e.

angular frequency = 2f = 12.57 rads/s

=k/m = 12.57, k = 31.6 N/m

When pulled by 2.5 cm to the right total displacement y = 2.5cm

PE of the spring = -kx²/2 = -31.6* (0.025)²/2= -0.009875 J

initial velocity v = 54.41 cm/s

KE = mv²/2 =0.2*0.544² /2 = 0.074 J

Total energy of the system 0.074-0.009875 = 0.064125 J

This will be equal to the maximum PE of the spring kx²/2 = 0.064125

x= 0.0637 m   = 6.37 cm

the motion is represented by, =k/m, is a constant of the spring and will not change

y = 6.37Sin(12.57t+)

when t=0, y = 2.5 cm, Sin() = 1, phaseshit = /2

Period T = 1/f = ½ = 0.5 s

=k/m = 12.57,   spring constant k = 31.60 N/m

x = -6.37Sin(12.5t+/2)

amplitude A= 6.37cm

speed v = dy/dt =6.37*12.5Sin(12.5t)

maximum speed v = A = 6.37*12.5 = 79.625 cm/s

maximum acceleration a = A² = 6.37*12.5² = 995.3125 cm/s

x(t) = -6.37Cos(12.57t),

v(t) = dx/dt = 79.625Sin(12.5t)

a(t) = -995.3125 Cos(12.5t)