6.3 Solutions
263
Therefore, time lost in 43,200 oscillations
= 42,300 × 0.0001 = 4.32 s
6.24 g = g 0
1 −
d
R
(1)
where g and g 0 are the acceleration due to gravity at depth d and surface,
respectively, and R is the radius of the earth.
T = T 0
g 0
g
= T 0
1 −
d
R
−1/2
= T 0
1 +
d
2R
Time registered for the whole day will be proportional to the time period. Thus
T
T 0
=
t
t 0
= 1 +
d
2R
86,400
86,400 − 300
= 1 +
d
2R
Substituting R = 6400 km, we find d = 44.6 km.
6.25 (a) Let the liquid level in the left limb be depressed by x, so that it is elevated
by the same height in the right limb (Fig. 6.17). If ρ is the density of the
liquid, A the cross-section of the tube, M the total mass, and m the mass
of liquid corresponding to the length 2x, which provides the unbalanced
force,
Md 2 x
dt 2 = −mg = −(2x Aρ)g
d 2 x
dt 2 = −
2Aρg
M
x = −
2Aρgx
h Aρ
= −
2gx
h
= −ω
2 x
This is the equation of SHM.
Fig. 6.17
263
Therefore, time lost in 43,200 oscillations
= 42,300 × 0.0001 = 4.32 s
6.24 g = g 0
1 −
d
R
(1)
where g and g 0 are the acceleration due to gravity at depth d and surface,
respectively, and R is the radius of the earth.
T = T 0
g 0
g
= T 0
1 −
d
R
−1/2
= T 0
1 +
d
2R
Time registered for the whole day will be proportional to the time period. Thus
T
T 0
=
t
t 0
= 1 +
d
2R
86,400
86,400 − 300
= 1 +
d
2R
Substituting R = 6400 km, we find d = 44.6 km.
6.25 (a) Let the liquid level in the left limb be depressed by x, so that it is elevated
by the same height in the right limb (Fig. 6.17). If ρ is the density of the
liquid, A the cross-section of the tube, M the total mass, and m the mass
of liquid corresponding to the length 2x, which provides the unbalanced
force,
Md 2 x
dt 2 = −mg = −(2x Aρ)g
d 2 x
dt 2 = −
2Aρg
M
x = −
2Aρgx
h Aρ
= −
2gx
h
= −ω
2 x
This is the equation of SHM.
Fig. 6.17