Physics Help Forum Amplitude of the oscillation

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Jan 29th 2014, 07:12 PM #1 Junior Member   Join Date: Jan 2014 Location: Indonesia Posts: 8 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...
 Jan 30th 2014, 12:58 AM #2 Junior Member   Join Date: Jan 2014 Posts: 15 I think that the solve is:
 Jan 30th 2014, 02:28 PM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,331 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).
 Feb 1st 2014, 06:29 AM #4 Junior Member   Join Date: Jan 2014 Location: Indonesia Posts: 8 Thanks @panda and @chipB. I got it

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