http://www.ck12.org Chapter 10. Periodic Motion
FIGURE 10.7
of the block isxand the final velocity of the block isv, the final energy of the block-spring system can be written as
Ef=^12 kx^2 +^12 mv^2.
Solving, we haveEi=Ef
→
1
2
kA^2 =1
2
kx^2 +1
2
mv^2 →1
2
( 75. 0 )( 0. 35 )^2 =
1
2
( 75. 0 )( 0. 15 )^2 +
1
2
( 5. 00 )v^2 →5. 00 v^2 = ( 75. 0 )( 0. 35 )^2 −( 75. 0 )( 0. 15 )^2 →v^2 =( 75. 0 )( 0. 35 )^2 −( 75. 0 )( 0. 15 )^2
5. 00
→
v^2 = 0. 100 →v=± 0. 316 →± 0. 32m
sNotice there are two solutions to the problem. Since the block is moving toward the left, the correct solution is
v=− 0. 32 ms.
b. What is the block’s speed when it passes through its equilibrium position?
Answer:
This occurs at the position where the spring is no longer stretched, that is, when it is at equilibrium. Taking the initial
energy at the block’s maximum extension (its amplitudeA= 0. 35 m) and the final energy at equilibriumx= 0. 00 m,
the conservation of mechanical energyEi=Efgives
1
2
kA^2 =1
2
kx^2 +1
2
mv^2 →1
2
( 75 )( 0. 35 )^2 =
1
2
( 75. 0 )( 0 )^2 +
1
2
( 5. 00 )v^2 →v^2 =75. 0 ( 0. 35 )^2
5. 00
→v=± 1. 36 .The speed of the block is therefore 1. 4
m
s.
c. What is the period of oscillationsTof the block?
Answer:T= 2 π
√m
k=^2 π√
5. 00
75. 0 =^1.^622 →^1.^62 sd. How much time does it take the block to go from its maximum extension of 0.35 m to its equilibrium position
0.00 m?
Answer;We will solve the problem in two different ways:
i. Since SHM is the projection of circular motion, the time for an object to travel from its maximum extension to