The amplitude of a particle executing SHM is 3 cm.The displacement at ...
The amplitude of a particle executing SHM is 3 cm.The displacement at ...
To find the displacement at which the kinetic energy is 25% more than the potential energy, we can start by understanding the relationship between kinetic energy (KE) and potential energy (PE) in simple harmonic motion (SHM).
In SHM, the total mechanical energy (E) remains constant throughout the motion. This means that the sum of kinetic energy and potential energy at any point in time is constant.
E = KE + PE
Since the amplitude of the particle executing SHM is given as 3 cm, we can express the potential energy and kinetic energy in terms of the displacement (x) from the equilibrium position.
Potential Energy (PE):
The potential energy at any point in SHM is given by the equation:
PE = (1/2)kx^2
where k is the force constant or the spring constant.
Kinetic Energy (KE):
The kinetic energy at any point in SHM is given by the equation:
KE = (1/2)mv^2
where m is the mass of the particle and v is its velocity.
Finding the displacement:
To find the displacement at which the kinetic energy is 25% more than the potential energy, we need to set up the equation:
KE = 1.25 * PE
Substituting the equations for KE and PE:
(1/2)mv^2 = 1.25 * (1/2)kx^2
Canceling the common factors and simplifying:
mv^2 = 1.25 * kx^2
Since v = ω√(A^2 - x^2), where ω is the angular frequency and A is the amplitude, we can substitute this value in the equation:
m(ω√(A^2 - x^2))^2 = 1.25 * kx^2
Simplifying further:
mω^2(A^2 - x^2) = 1.25kx^2
mω^2A^2 - mω^2x^2 = 1.25kx^2
mω^2A^2 = (1.25k + mω^2)x^2
Dividing both sides by (1.25k + mω^2):
(mω^2A^2)/(1.25k + mω^2) = x^2
Taking the square root of both sides:
x = √((mω^2A^2)/(1.25k + mω^2))
Since we know the amplitude A, we can substitute the given values into this equation to find the displacement at which the kinetic energy is 25% more than the potential energy.
x = √((mω^2(3)^2)/(1.25k + mω^2))
However, without specific values for mass (m), angular frequency (ω), and spring constant (k), we cannot calculate the exact displacement. Hence, we cannot determine the specific value for the displacement.
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