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Question

A particle is executing SHM along a straight line. Its velocities at distances 2 cm and 4 cm from the mean position are 3 cm/s and 2 cm/s respectively. Find the time period of SHM.

A
2π125 s
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B
π125 s
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C
2π65 s
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D
π85 s
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Solution

The correct option is A 2π125 s
Given,
x1=2 cm;x2=4 cm
v1=2 cm/s;v2=3 cm/s
Since, the particle is executing SHM. we use
v2=ω2(A2x2)
From which,
v21=ω2(A2x21) ... (1)
v22=ω2(A2x22) ... (2)
On subtracting (1) and (2)
v21v22=ω2(x22x21)
ω= v21v22x22x21
We know that,
Time period (T)=2πω
=2π x22x21v21v22
=2×π×(4)2(2)2(3)2(2)2
=2×π×125

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