Question

Standard Exe Mass On A block with mass m - 7.2 kg is attached to two springs with spring constants kleft -29 N/m and Kright-51 N/m. The block is pulled a distance x 0.24 m to the left of its equilibrium position and released from rest. 1) What is the magnitude of the net force on the block (the moment it is released)? 3.34 Submit 2) What is the effective spring constant of the two springs? N/m Submit 3) What is the period of oscillation of the block? Submit 4) How long does it take the block to return to equilibrium for the first time? Submit 5) What is the speed of the block as it passes through the equilibrium position? 6) What is the magnitude of the acceleration of the block as it passes through equilibrium? /s2 Submit 7) Where is the block located, relative to equilibrium, at a time 1.04 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value) Submit 7) where is the block located, relative to equilibrium, at a time 1.04 s after it is released?(if the block ts left of equilibrium give the answer as a negative value; if the block is right of equilbrium give the answer as a positive value) Submit Your submissions:.084 Computed value: .084 Subnitted: Friday, May 10 at 4:56 AM 8) What is the net force on the block at this time 1.04 s? (a negative force is to the left; a positive force is to the right) N Subnit 9) What is the total energy stored in the system? Subnit 10) If the block had been given an initial push, how would the period of oscillation change? the period would increase the period would decrease the period would not change Submit Survey Question) 11) Below is some space to write notes on this problerm

Answer 1

The case is related to ( Two Springs and a mass )

The equation of motion of the system becomes

mX = -k1x-k2x = -( k1 + k2 )x ( X= double differentiation of displacement ) ( The negative sign is taken as in the above case if one spring compress the other get streached form the mean position)

solving the equation we get

mX + ((K1+K2)/m)*x = 0

1) Net force

The net force will be sum of restoring forces acting on the block ( The restoring force that will act on by spring on left will be K(left)* m and that by the right will be k(right)*m

Net force = K(left)* m+ k(right)*m = 29*0.72+ 51*0.72 = 57.6N

2)

Effective Spring Constant = k1+k2 = 29+51 = 70

3)

w = (1/ 2*3.14) Root (( K1+ K2 ) /m ) = 0.15* (Root (( 29+51) / 7.2))= 0.49 ( w= oscillation frequency )