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A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle is
- a)5
- b)15
- c)45
- d)20
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A particle of mass of 4 kg suspended from a spring of force constant ...
Here,m = 4 k g ;k = 800 N m-1; E = 4J
In SHM , E = 1/2 kA2
Maximum acceleration, amax = ω2A
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A particle of mass of 4 kg suspended from a spring of force constant ...
Given Data:
- Mass of the particle (m) = 4 kg
- Force constant of the spring (k) = 800 N m^–1
- Total energy of the oscillator (E) = 4 J
Formula:
- Total energy of the oscillator (E) = 1/2 * k * A^2 (where A is the amplitude of oscillation)
- Maximum acceleration (a) = k * A / m
Solution:
- From the formula of total energy, we can find the amplitude of oscillation:
4 = 1/2 * 800 * A^2
A^2 = 4 / 400
A = 0.1 m
- Now, we can find the maximum acceleration using the formula:
a = 800 * 0.1 / 4
a = 20 m/s^2
Therefore, the maximum acceleration of the particle is 20 m/s^2, which corresponds to option 'D'.
- Mass of the particle (m) = 4 kg
- Force constant of the spring (k) = 800 N m^–1
- Total energy of the oscillator (E) = 4 J
Formula:
- Total energy of the oscillator (E) = 1/2 * k * A^2 (where A is the amplitude of oscillation)
- Maximum acceleration (a) = k * A / m
Solution:
- From the formula of total energy, we can find the amplitude of oscillation:
4 = 1/2 * 800 * A^2
A^2 = 4 / 400
A = 0.1 m
- Now, we can find the maximum acceleration using the formula:
a = 800 * 0.1 / 4
a = 20 m/s^2
Therefore, the maximum acceleration of the particle is 20 m/s^2, which corresponds to option 'D'.
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A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle isa)5b)15c)45d)20Correct answer is option 'D'. Can you explain this answer?
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A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle isa)5b)15c)45d)20Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle isa)5b)15c)45d)20Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle isa)5b)15c)45d)20Correct answer is option 'D'. Can you explain this answer?.
A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle isa)5b)15c)45d)20Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle isa)5b)15c)45d)20Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass of 4 kg suspended from a spring of force constant 800 N m–1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in m s–2) of the particle isa)5b)15c)45d)20Correct answer is option 'D'. Can you explain this answer?.
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