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JEE Advanced Model Paper 10 with solutions for online practice
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© examsnet.com
Question : 10
Total: 54
A particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passesthrough a certain point P (AP < BP) at successive interval of 0.5 seconds and 1.5 seconds with speed 3 m/s.Then
Maximum speed of particle is
3
√
2
m/s
Maximum speed of particle is
4
√
2
m/s
The ratio,
A
P
B
P
=
√
2
−
1
√
2
+
1
The ratio
A
P
B
P
=
1
√
2
Validate
Solution:
Let x = A sin (ωt +
ϕ
)
Putting the required conditions
ϕ
= 45°
v
m
a
x
=
3
√
2
m/s
A
P
B
P
=
1
−
1
√
2
1
+
1
√
2
=
√
2
−
1
√
2
+
1
© examsnet.com
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