A simple harmonic oscillator consists of a block of mass 4.10 kg attached to a spring of spring constant 300 N/m. When t = 1.80 s, the position and velocity of the block are x = 0.188 m and v = 4.150 m/s. What is the amplitude of the oscillations?

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Answer 1 · 2015-10-09T22:04:29

#A=0.7"m"#

The time period of the spring is given by:

#T=2pisqrt((m)/(k)) " "color(red)((1))#

The displacement #y# is given by:

#y=Asin2pift " "color(red)((2))#

#A# is the amplitude

#f# is the frequency

#t# is the time passed

#f=1/T " "color(red)((3))#

Substituting this into #color(red)((2))# for #f# gives:

#y=Asin(2pi1/Tt)#

Substituting #T# from #color(red)((1))# gives:

#y=Asin(cancel(2pi)1/cancel(2pi).sqrt((k)/(m)).t)rArr#

#y=Asin(sqrt((k)/(m)).t)#

#0.188=Asin[sqrt((300)/(4.1))xx1.8]#

#0.188=Asin(15.397)=Axx0.265#

#A=0.188/0.265=0.7"m"#

Answer 2 · 2015-10-09T23:44:28

..mind you...my maths can be wrong!

Have a look:
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