Chapter 10

Answer to p
roblems
– 27, 29, 32, 43, 47, 48 , 62, 63
27.
.
The period of oscillation of a spring

and

mass system is 0.50 s and the amplitude is 5.0 cm. What is the
magnitude of the acceleration at the point of maximum extension of the spring?
Give
n: A=5.0cm=5.0x10
2
m, T=0.50s, Question: a at maximum extension ?
At the maximum magnitude of the acceleration x=A the amplitude, v
m
=
A, am=
2
A
But
= 2p / T
a
m
ax
= A (2p/T)
2
=(5.0x10
2
m)x[4 p
2
/ (0.50s)
2
] = 7.9 m/s
2
43
.
An ideal spring with a spring constant of 15 N/m is suspended vertically. A body of mass 0.60 kg is
attached to the unstretched spring and released. (a) W
hat is the extension of the spring when the speed
is a maximum? (b) What is the maximum speed?
Given: m=0.60kg, Question: a) x at v
max
? b) v
max
?
It is at the equilibrium point that the speed is maximum and it is minimum when the
Amp
litude is reached. The forces are applied in the y direction
Writing newton’s law at equilibrium,
SF
y
=0
a)
SF
y
= kx
–
mg = 0
x = mg / k =( 0.60kg x 9.80 m/s
2
)
/15N = 0.39 m
b)
v
m
=
A = x v(k/m) = 0.39m v(15N/0.60kg) = 1.95 m/s
48.
.
A body is suspended vertically from an ideal spring of spring constant 2.5 N/m. The spring is initially
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 Spring '09
 HUANG
 amax, maximum kinetic energy, gravitational field strength

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