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Scelta multipla
At what displacement (expressed as a fraction of the amplitude A of the motion) is the kinetic energy of the cart in simple harmonic motion half of its maximum value?
A
A/2
B
A/√2
C
A/3
D
A/4
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1
Start by recalling the relationship between kinetic energy (KE) and potential energy (PE) in simple harmonic motion. The total mechanical energy (E) is constant and is the sum of KE and PE.
The maximum kinetic energy occurs when the potential energy is zero, which is at the equilibrium position. At this point, KE_max = (1/2)mv^2_max, where m is the mass and v_max is the maximum velocity.
The kinetic energy at any displacement x is given by KE = (1/2)mv^2 = (1/2)k(A^2 - x^2), where k is the spring constant and A is the amplitude.
Set the kinetic energy equal to half of its maximum value: (1/2)k(A^2 - x^2) = (1/2) * (1/2)kA^2. Simplify this equation to find the relationship between x and A.
Solve the equation for x in terms of A: A^2 - x^2 = (1/2)A^2. This simplifies to x^2 = (1/2)A^2, leading to x = A/√2. Thus, the displacement is A/√2.