(III) We usually neglect the mass of a spring if it is small compared to the truss attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length l and mass M S uniformly distributed along the length of the spring. A mass m is attached to the end of the spring. One end of the spring is fixed and the mass m is allowed to vibrate horizontally without friction (Fig. 7–30). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed ϑ 0 , the midpoint of the spring moves with speed ϑ 0 / 2 Show that the kinetic energy of the mass plus spring when the mass is moving with velocity ϑ is K = 1 2 M ϑ 2 where M = m 1 3 M S is the “effective mass” of the system. [ Hint : Let D be the total length of the stretched spring. Then the velocity of a mass d m of a spring of length d x located at x is ϑ ( x ) − ϑ 0 ( x / D ) . Note also d m = d x ( M S / D ) .] FIGURE 7-30 Problem 68.
(III) We usually neglect the mass of a spring if it is small compared to the truss attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length l and mass M S uniformly distributed along the length of the spring. A mass m is attached to the end of the spring. One end of the spring is fixed and the mass m is allowed to vibrate horizontally without friction (Fig. 7–30). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed ϑ 0 , the midpoint of the spring moves with speed ϑ 0 / 2 Show that the kinetic energy of the mass plus spring when the mass is moving with velocity ϑ is K = 1 2 M ϑ 2 where M = m 1 3 M S is the “effective mass” of the system. [ Hint : Let D be the total length of the stretched spring. Then the velocity of a mass d m of a spring of length d x located at x is ϑ ( x ) − ϑ 0 ( x / D ) . Note also d m = d x ( M S / D ) .] FIGURE 7-30 Problem 68.
(III) We usually neglect the mass of a spring if it is small compared to the truss attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length
l
and mass
M
S
uniformly distributed along the length of the spring. A mass
m
is attached to the end of the spring. One end of the spring is fixed and the mass m is allowed to vibrate horizontally without friction (Fig. 7–30). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed
ϑ
0
, the midpoint of the spring moves with speed
ϑ
0
/
2
Show that the kinetic energy of the mass plus spring when the mass is moving with velocity
ϑ
is
K
=
1
2
M
ϑ
2
where
M
=
m
1
3
M
S
is the “effective mass” of the system. [Hint: Let D be the total length of the stretched spring. Then the velocity of a mass
d
m
of a spring of length
d
x
located at
x
is
ϑ
(
x
)
−
ϑ
0
(
x
/
D
)
. Note also
d
m
=
d
x
(
M
S
/
D
)
.]
(III) A clock pendulum oscillates at a frequency of 2.5 Hz.At t=0 it is released from rest starting at an angle of 12°to the vertical. Ignoring friction, what will be the position(angle in radians) of the pendulum at (a) t=0.25s(b) t=1.60s and (c) t=500s
(II) A mass of 240 g oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm. (a) What is the effective spring constant for this motion? (b) How much energy is involved in this motion?
(I) How long must a simple pendulum be if it is to make exactly one swing per second? (That is, one complete oscillation takes exactly 2.0 s.)
Chapter 7 Solutions
Physics for Scientists and Engineers with Modern Physics
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