A vibrator of mass 1gm is acted upon by a restoring force per unit displacement of 10^(4)Nm^(-1) retarding force per unit velocity of 4Nm^(-1) sec and a driving force of cospt N . Calculate the value of maximum possible amplitude.
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Answr;
A=2‾√cm
A
=
2
c
m
,n=1000πHz
n
=
1000
π
H
z
A=12‾√cm
A
=
1
2
c
m
,n=500πHz
n
=
500
π
H
z
A=12‾√cm
A
=
1
2
c
m
,n=1000π
n
=
1000
π
Hz
Answer :
A
Solution :
For a displacement x, the kinetic and potential energies are KE=12m(A2−x2)ω2
K
E
=
1
2
m
(
A
2
-
x
2
)
ω
2
and PE=12mx2ω2
P
E
=
1
2
m
x
2
ω
2
Each of these =102=5J
=
10
2
=
5
J
when x=1cm
x
=
1
c
m
Hence 12mx2ω2=12×0.1×(1×10−2)2ω2=5
1
2
m
x
2
ω
2
=
1
2
×
0.1
×
(
1
×
10
-
2
)
2
ω
2
=
5
This gives ω2=100.1×10−4=106
ω
2
=
10
0.1
×
10
-
4
=
10
6
Giving ω=103=1000rads
ω
=
10
3
=
1000
r
a
d
s
Hence T=2πω=π500s
T
=
2
π
ω
=
π
500
s
And frequency f=500πs
f
=
500
π
s
Also, KE=5=12×0.1[A2−(1×10−2)2](106)
K
E
=
5
=
1
2
×
0.1
[
A
2
-
(
1
×
10
-
2
)
2
]
(
10
6
)
A2=2×10−4⇒A=2‾√×10−2m=2‾√cm
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