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Simple harmonic motion question 
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#1
Jul210, 01:02 AM

P: 6

1. The problem statement, all variables and given/known data
A block of unknown mass is attached to a spring of spring constant 6.7N/m and undergoes simple harmonic motion with an amplitude of 7.8 cm. When the mass is halfway between its equilibrium position and its max endpoint, its speed is measured to be 38.5 cm/s. (a) Calculate the mass of the block. (b) Find the period of the motion. (c) Calculate the maximum acceleration. 2. Relevant equations 1. E = K + U_{s} 2. E= 1/2 kA^{2} 3. omega = [tex]\sqrt{k/m}[/tex] 4. T = 2pi/w 5. x(t) = Acos(omega*t+phi) 6. v(t) = Awsin(omega]t+phi)(not sure if this is right...) 7. a(t) = Aw^{2}cos(wt+phi)(not really sure if I took the derivatives correctly...) 3. The attempt at a solution I was only given the x(t) function for equations 6&7 so please correct me if I'm wrong on those. But here is what I have done so far...(lower case k is spring constant. upper case K is kinetic energy) (a) 1/2kA^{2} = 1/2kx^{2} + 1/2mv^{2} 407.63 = 101.94 + 38.5 m 305.69/38.5 = m m = 7.94 kg (should it be kg? right answer?) (b) w = sqrt(7.8 / 7.94) = 0.99 T = 2pi / 0.99 = 6.34 (right answer?) (c) (this is where i really need help) I don't know how to find a_{max}. Do i set phi equal to a certain angle (radians)? Here is what I have so far, but not really sure if I'm doing it right. a(t) = Aw^{2}cos(wt +phi) = (7.8)(0.99)^{2}cos{(0.99)(38.5)+phi} I just don't know what to set phi equal to or what I need to do from here (assuming I took the derivatives of x(t) correctly.) I really do appreciate all of your help and hard work to make this an awesome help site! Much love, Tim 


#2
Jul210, 01:19 AM

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#3
Jul210, 01:31 AM

P: 6

So i guess i should change everything to m? 


#4
Jul210, 02:41 AM

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Simple harmonic motion question



#5
Jul210, 02:51 AM

P: 6




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