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A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy half potential and half kinetic?
- a)10 cm
- b)√2 cm
- c)2 cm
- d)2√2 cm
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A particle is vibrating in simple harmonic motion with an amplitude o...
The total energy E of a particle vibrating SHM is given by
The kinetic energy K is given by
where y = displacements of the particle
But, 
Hence the kinetic energy is half of the total energy when displacement of the particle is a/√2. Given that a = 4cm.
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Community Answer
A particle is vibrating in simple harmonic motion with an amplitude o...
Approach:
To find the displacement from the equilibrium position where the energy is half potential and half kinetic, we need to consider the total mechanical energy of the particle in simple harmonic motion.
Given:
Amplitude (A) = 4 cm
Formula:
Total mechanical energy (E) = Potential energy (PE) + Kinetic energy (KE)
E = 1/2 kA^2
where k is the spring constant.
Solution:
1. We know that at any displacement x from the equilibrium point, the total mechanical energy is given by:
E = 1/2 kx^2
2. At the point where the energy is half potential and half kinetic, we have:
1/2 kx^2 = 1/2 kA^2
x^2 = A^2
x = A = 4 cm
3. Therefore, the displacement from the equilibrium position where the energy is half potential and half kinetic is:
x = 4 cm = 4√2 cm
Final Answer:
The correct option is 2√2 cm (Option D).
To find the displacement from the equilibrium position where the energy is half potential and half kinetic, we need to consider the total mechanical energy of the particle in simple harmonic motion.
Given:
Amplitude (A) = 4 cm
Formula:
Total mechanical energy (E) = Potential energy (PE) + Kinetic energy (KE)
E = 1/2 kA^2
where k is the spring constant.
Solution:
1. We know that at any displacement x from the equilibrium point, the total mechanical energy is given by:
E = 1/2 kx^2
2. At the point where the energy is half potential and half kinetic, we have:
1/2 kx^2 = 1/2 kA^2
x^2 = A^2
x = A = 4 cm
3. Therefore, the displacement from the equilibrium position where the energy is half potential and half kinetic is:
x = 4 cm = 4√2 cm
Final Answer:
The correct option is 2√2 cm (Option D).
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A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy half potential and half kinetic?a)10 cmb)√2 cmc)2 cmd)2√2 cmCorrect answer is option 'D'. Can you explain this answer?
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