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Old Jan 29th 2014, 07:12 PM   #1
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Amplitude of the oscillation

Another question, this is about mass on helical spring

A mass is attached to the wall by a spring of constant k. When the spring is at its natural length, the mass is given a certain initial velocity, resulting in oscillations of Amplitude A. If the spring is replaced by a spring constant 2k and the mass is given the same initial velocity, what is the amplitude of the resulting oscillation?

Ans : a) A√2
b) 2 A√2
c) A√2 /2
d) (3/2) A√2
e) (2/3) A√2

using the formula I've found this : A' = A sin √2Ѡt

Please help...
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Old Jan 30th 2014, 12:58 AM   #2
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I think that the solve is:
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Old Jan 30th 2014, 02:28 PM   #3
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One way to approach this is with energy principles. With the spring in a neutral position (unstretched) the energy of the system is 1/2 mv^2, whereas with the spring fully extended v=0 and the energy of the system is 1/2 kx^2. So two different springs k_1 and k_2, given the same initial kinetic energy of the mass m, will have max extensions such that:

(1/2) k_1 x_1^2 = (1/2) k_2 x_2^2

Thus (x_2/x_1) = sqrt(k_1/k_2)

and if k_2 is two times k_1 this means that the ratio of amplitudes is:

x_2/x_1 = sqrt(1/2).
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Old Feb 1st 2014, 06:29 AM   #4
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Smile

Thanks @panda and @chipB. I got it
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