Suppose a spring with an attached mass of 2kg has a natural length 0.5m. A 25.6

Question

Suppose a spring with an attached mass of 2kg has a natural length 0.5m.
A 25.6 N force is required to stretch it to a length of 0.7m. If the spring is stretched to a length of 0.7m and then releases with initial velocity 0,
A.)find the position of the mass at any time t.
B.)Suppose the system is attached to a shock absorber(a type of damping device) with damping constant c=40. Find the position of the mass at time t, give that it has an initial velocity of 0.6m/s and it starts from the equilibrium position.

Explanation / Answer

The net displacement in the string generates a force kx

kx = 25.6N (given)

x=0.2

=> k*0.2 = 25.6

=> k = 128 N/m ;

once the force 25.6 N was removed....

the net force acting on the block is kx from the spring...

=> kx = ma = -2*(d2x/dt2); a is negative as x is decreasing...

=>2(d2x/dt2)+kx=0

=>2(d2x/dt2)+128x=0 => a = -2x with = 8

the general solution of the above equation is

x(t)=Acos8t+Bsin8t ;

x(0) = A = 0.2 ;

B = 0 after differentiation

=> x(t) = 0.2cos8t

2)

now given v(0) = 0.6 => B = 0.6/8 = 3/40 ;

=> x(t) = 0.2cos8t + 3/40sin8t ;

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