1000 Solved Problems in Modern Physics

(Romina) #1

1.3 Solutions 61


Fig. 1.16Two modes of Oscillation


If initiallyx 1 =x 2 , the masses oscillate in phase with frequencyω 1 (sym-
metrical mode) as in Fig. 1.16(a). If initiallyx 2 =−x 2 then the masses
oscillate out of phase (asymmetrical) as in Fig. 1.16(b)

1.62 Sum of translational+rotational+potential energy=constant
1
2


mv^2 +

1

2

Iω^2 +

1

2

kx^2 =const.

ButI=

1

2

mR^2 andω=v/R

Therefore

3

4

mv^2 +

1

2

kx^2 =const.
3
4

m(dx/dt)^2 +

1

2

kx^2 =const.
Differentiating with respect to time,
(
3
2

)(

md^2 x
dt^2

)

.

dx
dt

+kx.

dx
dt

= 0

Cancelling dx/dtthrough and simplifying d^2 x/dt^2 +(2k/ 3 m)x=0. This
is an equation to SHM.
Writingω^2 = 32 mk, time periodT=^2 ωπ= 2 π


3 m
2 k

1.63


d^2 y
dx^2

− 8

dy
dx

=− 16 y

d^2 y
dx^2

− 8

dy
dx

+ 16 y= 0

Auxiliary equation:
D^2 − 8 D+ 16 = 0
(D−4)(D−4)= 0
The roots are 4 and 4.
Thereforey=C 1 e^4 x+C 2 xe^4 x

1.64x^2


dy
dx

+y(x+1)x= 9 x^2 (1)

Put the above equation in the form
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