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A particle perform S.H.M of amplitude A along a straight line . when it is at a distance 3/2A from mean position , it's K.E gets increased by an amount U,(x)( T.M.E) due to an impulsive force Then it's New amplitude becomes,
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A particle perform S.H.M of amplitude A along a straight line . when i...
Explanation:
To solve the given problem, we need to understand the concept of Simple Harmonic Motion (S.H.M) and impulsive forces.
Simple Harmonic Motion (S.H.M):
S.H.M is a type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts towards the equilibrium position. The equation of S.H.M is given by:
x = A * sin(ωt + φ)
where,
x = displacement from the mean position
A = amplitude of the motion
ω = angular frequency
t = time
φ = phase constant
Impulsive Force:
An impulsive force is a force that acts on a body for a very short duration of time. It changes the momentum of the body instantaneously and causes a sudden change in its velocity.
Given:
Amplitude of S.H.M (A) = A
Distance from mean position (x) = 3/2A
Increase in K.E (ΔKE) = U * T.M.E
Solution:
1. Let's assume the initial position of the particle is at the mean position (x = 0).
2. At a distance of 3/2A from the mean position, the particle will have a displacement of x = 3/2A.
3. Using the equation of S.H.M, we can find the value of ωt + φ at x = 3/2A.
3/2A = A * sin(ωt + φ)
sin(ωt + φ) = 3/2
ωt + φ = sin^(-1)(3/2)
ωt + φ = π/2
ωt = π/2 - φ
4. Now, let's consider the impulsive force acting on the particle. Due to this force, the particle's K.E increases by an amount ΔKE = U * T.M.E.
ΔKE = U * T.M.E = U * (1/2) * (mω^2A^2)
5. The new amplitude of the particle's motion can be calculated using the relation between K.E and amplitude.
ΔKE = (1/2) * mω^2 * (A^2 - A'^2)
U * (1/2) * (mω^2A^2) = (1/2) * mω^2 * (A^2 - A'^2)
U * A^2 = A^2 - A'^2
A'^2 = A^2 - U * A^2
A' = A * √(1 - U)
Therefore, the new amplitude of the particle's motion is A * √(1 - U).
To solve the given problem, we need to understand the concept of Simple Harmonic Motion (S.H.M) and impulsive forces.
Simple Harmonic Motion (S.H.M):
S.H.M is a type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts towards the equilibrium position. The equation of S.H.M is given by:
x = A * sin(ωt + φ)
where,
x = displacement from the mean position
A = amplitude of the motion
ω = angular frequency
t = time
φ = phase constant
Impulsive Force:
An impulsive force is a force that acts on a body for a very short duration of time. It changes the momentum of the body instantaneously and causes a sudden change in its velocity.
Given:
Amplitude of S.H.M (A) = A
Distance from mean position (x) = 3/2A
Increase in K.E (ΔKE) = U * T.M.E
Solution:
1. Let's assume the initial position of the particle is at the mean position (x = 0).
2. At a distance of 3/2A from the mean position, the particle will have a displacement of x = 3/2A.
3. Using the equation of S.H.M, we can find the value of ωt + φ at x = 3/2A.
3/2A = A * sin(ωt + φ)
sin(ωt + φ) = 3/2
ωt + φ = sin^(-1)(3/2)
ωt + φ = π/2
ωt = π/2 - φ
4. Now, let's consider the impulsive force acting on the particle. Due to this force, the particle's K.E increases by an amount ΔKE = U * T.M.E.
ΔKE = U * T.M.E = U * (1/2) * (mω^2A^2)
5. The new amplitude of the particle's motion can be calculated using the relation between K.E and amplitude.
ΔKE = (1/2) * mω^2 * (A^2 - A'^2)
U * (1/2) * (mω^2A^2) = (1/2) * mω^2 * (A^2 - A'^2)
U * A^2 = A^2 - A'^2
A'^2 = A^2 - U * A^2
A' = A * √(1 - U)
Therefore, the new amplitude of the particle's motion is A * √(1 - U).
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A particle perform S.H.M of amplitude A along a straight line . when it is at a distance 3/2A from mean position , it's K.E gets increased by an amount U,(x)( T.M.E) due to an impulsive force Then it's New amplitude becomes, Related: Syllabus covered in each Part Test of JEE Advanced?
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