Dul De. 1028 20 1:59 00 PM End Date: 12/22/2017 8:00:00 PM (ii%) Problem 5: A ma

Question

Dul De. 1028 20 1:59 00 PM End Date: 12/22/2017 8:00:00 PM (ii%) Problem 5: A mass a spring of spring constant k is undergoing simple harmonic oscillations with amplitude m at the 33% Part (a) At what positive value of displacement x m terms of.d is the potential energy 1.9 of the total mechanical energy? 33% Part (b) what fraction of the total mechanical energy is kinetic if the displacement is l 2 the amplitude 33% Part (c) By what factor does the maximum kinetic energy change if the amplitude is increased by a factor of 3? 0% 100% Deductions

Explanation / Answer

using law of conservation of energy

total mechanical energy = maximum potential energy = 0.5*k*A^2

total energy at extreme position = total energy at the point where PE is (1/9)th of total mechanical energy

0.5*k*A^2 = 0.5*m*V^2 + (1/9)*0.5*k*A^2

but v is the velocity at the point where PE = (1/9)*TE

V^2 = w^2*(A^2-x^2)

then

0.5*k*A^2 = (0.5*m*w^2*(A^2-x^2)+(1/18)*k*A^2

(4/9)*k*A^2 = 0.5*m*w^2*(A^2-x^2)

but m*w^2 = spring constant k

then

(4/9)*k*A^2 = 0.5*k*(A^2-x^2)

k cancels

(4/9)*A^2 = 0.5*A^2 - 0.5*x^2

0.5*x^2 = ((1/2)-(4/9))*A^2 = (1/18)*A^2

(1/2)*x^2 = (1/18)*A^2

x = (1/3)*A

b)when x = (1/2)*A

PE = 0.5*k*x^2 = 0.5*k*(A^2/4)

Total energy is 0.5*k*A^2

then KE = 0.5*k*A^2 - (1/4)*0.5*k*A^2 = (3/4)*0.5*k*A^2

required fraction is KE/TE =(3/4)*0.5*k*A^2 / (0.5*k*A^2) = (3/4)

c) maximum KE =0.5*k*A^2

if amplitude is increased by 3 times then

KEmax is increases by (3^2) = 9 times

so answer is 9

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