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A body is executing S .H. M . When the displacements from the mean position are 4cm 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 [1991]
  • a)
    2π sec
  • b)
    π/2 sec
  • c)
    π sec
  • d)
    (3π/2)sec
Correct answer is option 'C'. Can you explain this answer?
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A body is executing S .H. M . When the displacements from the mean pos...
Understanding Simple Harmonic Motion (SHM)
In SHM, the relationship between displacement, velocity, and time period is crucial for solving problems.
Given Data
- Displacement 1 (x1) = 4 cm, Velocity 1 (v1) = 10 cm/s
- Displacement 2 (x2) = 5 cm, Velocity 2 (v2) = 8 cm/s
Key Equations
The formula for velocity in SHM is given by:
v = ω√(A² - x²)
Where:
- v = velocity
- ω = angular frequency
- A = amplitude
- x = displacement
Applying the Equations
1. For the first displacement:
10 = ω√(A² - 4²) (Equation 1)
2. For the second displacement:
8 = ω√(A² - 5²) (Equation 2)
Squaring the Equations
- From Equation 1:
100 = ω²(A² - 16)
- From Equation 2:
64 = ω²(A² - 25)
Subtracting the Equations
Subtract the second equation from the first:
100 - 64 = ω²[(A² - 16) - (A² - 25)]
36 = ω²(9)
ω² = 4
ω = 2 rad/s
Calculating Time Period (T)
The time period T is related to angular frequency by:
T = 2π/ω
Substituting the value of ω:
T = 2π/2 = π seconds
Conclusion
Thus, the time period of the body executing SHM is:
Correct Answer: (C) π seconds
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