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EAMCET Engineering 2003 Solved Paper
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© examsnet.com
Question : 8
Total: 160
When a body of mass
1.0
kg
is suspended from a certain light spring hanging vertically, its length increases by
5
cm
. By suspending
2.0
kg
block to the spring and if the block is pulled through
10
cm
and released, the maximum velocity in it in
m
∕
s
is : (Acceleration due to gravity
=
10
m
∕
s
2
)
0.5
1
2
4
Validate
Solution:
m
1
=
1
kg
, extension
l
1
=
5
cm
=
5
×
10
−
2
m
m
1
g
=
k
l
1
k
=
force constant of the spring
k
=
m
1
g
l
1
=
1
×
10
5
×
10
−
2
=
200
N
∕
m
Time period of the block of mass
2
kg
,
T
=
2
π
√
m
k
=
2
π
√
2
200
=
2
π
×
1
10
=
π
5
s
Maximum velocity,
v
max
=
A
ω
where,
A
=
amplitude
=
10
cm
=
10
×
10
−
2
m
v
max
=
A
×
2
π
T
=
10
×
10
−
2
×
2
π
π
∕
5
=
10
−
1
×
2
×
5
=
1
m
∕
s
© examsnet.com
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