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The maximum velocity of a particle executing simple harmonic is V. If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value, the maximum velocity becomes
- a)18V
- b)12V
- c)6V
- d)3V
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The maximum velocity of a particle executing simple harmonic is V. If ...
The maximum velocity of a particle executing simple harmonic motion is given by the equation V = ωA, where V is the maximum velocity, ω is the angular frequency, and A is the amplitude of the motion.
Now, let's consider the given scenario where the amplitude is doubled and the time period of oscillation is decreased to 1/3 of its original value.
Doubling the Amplitude
When the amplitude is doubled, the new amplitude becomes 2A. This means that the particle now oscillates between -2A and +2A instead of between -A and +A.
Decreasing the Time Period
When the time period is decreased to 1/3 of its original value, the new time period becomes T/3, where T is the original time period.
The Angular Frequency
The angular frequency is given by the equation ω = 2π/T, where T is the time period.
Since we know that the time period is decreased to 1/3 of its original value, the new angular frequency becomes ω' = 2π/(T/3) = 6π/T.
Calculating the Maximum Velocity
Using the equation V = ωA, we can calculate the maximum velocity for the new scenario:
V' = ω'A'
Substituting the new values of ω' and A', we get:
V' = (6π/T)(2A)
Simplifying, we get:
V' = 12πA/T
Since ω = 2π/T, we can rewrite the equation as:
V' = 6ωA
But we know that V = ωA, so the maximum velocity in the new scenario is six times the maximum velocity in the original scenario:
V' = 6V
Hence, the correct answer is option C: The maximum velocity becomes 6V.
Now, let's consider the given scenario where the amplitude is doubled and the time period of oscillation is decreased to 1/3 of its original value.
Doubling the Amplitude
When the amplitude is doubled, the new amplitude becomes 2A. This means that the particle now oscillates between -2A and +2A instead of between -A and +A.
Decreasing the Time Period
When the time period is decreased to 1/3 of its original value, the new time period becomes T/3, where T is the original time period.
The Angular Frequency
The angular frequency is given by the equation ω = 2π/T, where T is the time period.
Since we know that the time period is decreased to 1/3 of its original value, the new angular frequency becomes ω' = 2π/(T/3) = 6π/T.
Calculating the Maximum Velocity
Using the equation V = ωA, we can calculate the maximum velocity for the new scenario:
V' = ω'A'
Substituting the new values of ω' and A', we get:
V' = (6π/T)(2A)
Simplifying, we get:
V' = 12πA/T
Since ω = 2π/T, we can rewrite the equation as:
V' = 6ωA
But we know that V = ωA, so the maximum velocity in the new scenario is six times the maximum velocity in the original scenario:
V' = 6V
Hence, the correct answer is option C: The maximum velocity becomes 6V.
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Community Answer
The maximum velocity of a particle executing simple harmonic is V. If ...
V1 = A1W1
W1 =2π/T1
A2 = 2A1
T2 = T1/3
W2 =2π/T2 = 3(2π/T1)=3W1
V2 = A2W2 = 2A1(3W1) =6V1
W1 =2π/T1
A2 = 2A1
T2 = T1/3
W2 =2π/T2 = 3(2π/T1)=3W1
V2 = A2W2 = 2A1(3W1) =6V1
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The maximum velocity of a particle executing simple harmonic is V. If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value, the maximum velocity becomesa)18Vb)12Vc)6Vd)3VCorrect answer is option 'C'. Can you explain this answer?
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