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The magnitude of the maximum acceleration is π times that of the maximum velocity of a simple harmonic oscillator. The time period of the oscillator in seconds is
- a)4
- b)2
- c)1
- d)0.5
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The magnitude of the maximum acceleration is π times that of the maxi...
Maximum acceleration = Maximum velocity × π
I.e., ω2A = πωA
where A is amplitude and ω is angular velocity.
⇒ ω = π
⇒ 2π/T = π
⇒ T = 2 s
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The magnitude of the maximum acceleration is π times that of the maxi...
Explanation:
Given:
The magnitude of the maximum acceleration is π times that of the maximum velocity of a simple harmonic oscillator.
To find:
The time period of the oscillator in seconds.
Solution:
Relationship between acceleration and velocity:
In a simple harmonic oscillator, the maximum acceleration (a) is related to the maximum velocity (v) by the equation a = ω^2 * x, where ω is the angular frequency and x is the amplitude of the oscillator.
Given:
a = πv
a = ω^2 * x
Substitute a = πv into the equation:
πv = ω^2 * x
Relationship between angular frequency and time period:
The angular frequency (ω) is related to the time period (T) by the equation ω = 2π / T.
Substitute ω = 2π / T into the equation πv = ω^2 * x:
πv = (2π / T)^2 * x
πv = 4π^2x / T^2
Since v = ωx:
πωx = 4π^2x / T^2
T^2 = 4
T = 2 seconds
Therefore, the time period of the oscillator is 2 seconds (option B).
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The magnitude of the maximum acceleration is π times that of the maximum velocity of a simple harmonic oscillator. The time period of the oscillator in seconds isa)4b)2c)1d)0.5Correct answer is option 'B'. Can you explain this answer?
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