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An object of mass 0.35kg oscillates in SHM with an amplitude of 140mm

and a frequency of 0.60Hz.
Calculate or find:
i) Maximum kinetic energy of the object.
ii) Maximum potential energy of the object.
iii) Potential and kinetic energy at the mid-way point between the centre and the
extremity of the motion.

User Mamadrood
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8.3k points

1 Answer

5 votes

Answer:

The equation for the total mechanical energy of an object in SHM is:

E = 1/2 kA^2

where E is the total mechanical energy, k is the spring constant, and A is the amplitude.

To solve the problem, we need to find the spring constant of the oscillator:

f = 1/T

where f is the frequency and T is the period.

T = 1/f = 1/0.60 = 1.67 s

The angular frequency of the oscillator is:

ω = 2πf = 2π/T = 3.76 rad/s

The spring constant of the oscillator is:

k = mω^2 = 0.35 x (3.76)^2 = 4.97 N/m

i) The maximum kinetic energy of the object is equal to the maximum potential energy, which is:

Emax = 1/2 kA^2 = 1/2 x 4.97 x (0.14)^2 = 0.012 J

ii) The maximum potential energy of the object is the same as the maximum kinetic energy, which is:

Emax = 1/2 kA^2 = 1/2 x 4.97 x (0.14)^2 = 0.012 J

iii) At the mid-way point between the centre and the extremity of the motion, the displacement of the oscillator is half the amplitude, which is 70 mm or 0.07 m. At this point, the kinetic energy is zero, and the potential energy is:

E = 1/2 kx^2 = 1/2 x 4.97 x (0.07)^2 = 0.012 J

Therefore, the total mechanical energy at this point is also 0.012 J.

User Riccardo La Marca
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8.0k points