Energy in harmonic Motion (Basic) QUE STION 1 The total mechanical energy of a h

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Energy in harmonic Motion (Basic)

QUE STION 1 The total mechanical energy of a harmonic oscillator that consists of a spring with spring constant k and mass m is pring constant k and is the kinetic energy of the mass and Uelx" is the elastic potential energy of the spring. If there are no other dissipative forces acting (e.g. friction or airdrag are negligible), the total mechanical energy is conserved. This means the oscillator reaches maximum displacement xmax=A (amplitude) for v=0, and maximum speed vmax for x=0 A. Consider a harmonic oscillator with mass 0.26 kg and spring constant 178 N/m. If the speed of the mass at the equilibrium point is 1.83 m/s, what is the amplitude? (work this out using conservation of energy. Practise how to draw an energy bar chart for this.) Convert your Convert your answer to units of cm B. Consider a harmonic oscillator with mass 0.22 kg and spring constant 165 N/m. If the amplitude is 8.58 cm, what is the speed of the mass at the equilibrium point? (work this out using conservation of energy. Practise how to draw an energy bar chart for this.) for this) amplitude is 8 the equilbfium energy bar Answer with units of mls.

Explanation / Answer

A.)
m = 0.26 kg
k = 178 N/m

At equilibrium,
All the Energy is due to Kinetic Energy
Etot = 1/2 * m *v^2

Now At Max Compression/Contraction , All the Energy is due to Elastic Potential Energy,Therefore
1/2 * m *v^2 = 1/2 * kx^2
0.26 * 1.83^2 = 178 * x^2
x = 0.0699 m
Amplitude = 6.7 cm

B.)
m = 0.22 Kg
k = 165 N/m
X = 8.58 * 10^-2 mEtot

Using Same Energy Conservation Equation as above
1/2 * mv^2 = 1/2 * kx^2
0.22 * v^2 = 165 * (8.58 * 10^-2)^2
v = 2.35 m/s

Speed of the mass at equilibrium point, v = 2.35 m/s

(c)
w = sqrt(k/m)
k = w^2 * m
k = (2*pi*f)^2 * m

Using Same Energy Conservation Equation as above
1/2 * mv^2 = 1/2 * kx^2
1/2 * m*v^2 = 1/2 * (2*pi*f)^2 * m * x^2
v^2 = (2*pi*4.13)^2 * (8.0*10^-2)^2
v = 2.076 m/s

Speed of the mass at equilibrium point, v = 2.076 m/s

(D)
m = 0.47 Kg
k = 182 N/m
X = 8.01 * 10^-2 m

Total Energy = 1/2 * k*x^2
Etot = 1/2 * 182 * (8.01 * 10^-2)^2 J
Etot = 0.5838 J

Displacement , 0.58 * 8.01 cm = 4.65 cm
At this point
Potential Energy = 1/2*k*x^2
Potential Energy = 1/2 * 182 * (4.65 * 10^-2)^2 J
Potential Energy = 0.196 J

Kinetic Energy at this point, = Etot - Potential Energy
1/2 * mv^2 = 0.5838 - 0.196 J
1/2 * 0.47 * v^2 = 0.5838 - 0.196
v = 1.28 m/s
Speed of Mass at this point, v = 1.28 m/s

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