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A body is executing S.H.M When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocities of the body are 10 cm per sec and 8 cm per sec. Then the time period of the body is
  • a)
    2 π/2 sec
  • b)
    π sec
  • c)
    3 π/2 sec
  • d)
    2 π sec
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A body is executing S.H.M When its displacement from the mean position...
v=ω√A2−x2
​⇒8=ω √A2−25​
&10=ω√A2−16
​⇒(8/10​)2= A2−25/ A2−16
​⇒16/25​= A2−25/ A2−16
​⇒16A2−256=25A2−625
⇒9A2=369
A2=41
⇒8=ω41−25
​⇒8=4ω
∴ω=2=T2π​
T=π sec​
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Most Upvoted Answer
A body is executing S.H.M When its displacement from the mean position...
Given Data:
- Displacement from mean position (x1) = 4 cm, (x2) = 5 cm
- Velocities at x1 (v1) = 10 cm/sec, at x2 (v2) = 8 cm/sec

Formula:
The velocity of a body in Simple Harmonic Motion (S.H.M) is given by v = ω√(A^2 - x^2), where:
- v is the velocity of the body
- ω is the angular frequency of the motion
- A is the amplitude of the motion
- x is the displacement of the body from the mean position

Calculations:
Given v1 = 10 cm/sec and x1 = 4 cm:
10 = ω√(A^2 - 4^2)
100 = ω^2(A^2 - 16) ... (1)
Given v2 = 8 cm/sec and x2 = 5 cm:
8 = ω√(A^2 - 5^2)
64 = ω^2(A^2 - 25) ... (2)
Dividing equation (1) by equation (2):
100/64 = (A^2 - 16)/(A^2 - 25)
25/16 = (A^2 - 16)/(A^2 - 25)
25(A^2 - 25) = 16(A^2 - 16)
25A^2 - 625 = 16A^2 - 256
9A^2 = 369
A^2 = 369/9
A^2 = 41
A ≈ √41
A ≈ 6.4 cm
The time period of a body in S.H.M is given by T = 2π/ω, where ω is the angular frequency.
T = 2π/ω
T = 2π/√(g/A)
T = 2π/√(9.8/6.4)
T = 2π/√1.531
T ≈ 2 sec
Therefore, the time period of the body is 2 seconds (option B).
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A body is executing S.H.M When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocities of the body are 10 cm per sec and 8 cm per sec. Then the time period of the body isa)2π/2secb)πsecc)3π/2secd)2π secCorrect answer is option 'B'. Can you explain this answer?
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