Yahoo Answers: Answers and Comments for A particle attached to a spring with k = 48 N/m is undergoing simple harmonic motion, and its position is? [Physics]
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From cheer4u_n92
enUS
Fri, 25 Nov 2011 20:23:14 +0000
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Yahoo Answers: Answers and Comments for A particle attached to a spring with k = 48 N/m is undergoing simple harmonic motion, and its position is? [Physics]
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https://answers.yahoo.com/question/index?qid=20111125202314AAZJQta
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From Stephanie: Let m = mass of the particle. Assuming Hooke&#...
https://answers.yahoo.com/question/index?qid=20111125202314AAZJQta
https://answers.yahoo.com/question/index?qid=20111125202314AAZJQta
Mon, 16 May 2016 13:14:57 +0000
Let m = mass of the particle. Assuming Hooke's law holds, we have m x'' = k x, so x'' =  (k / m) x. Thus, x(t) = A cos (B t), and therefore B^2 = (k / m). On the other hand, we know that B = 7.0. It follows that m = k / 49.

From Yolanda: We are given the spring constant k=48 N/m, the...
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Fri, 25 Nov 2011 21:26:41 +0000
We are given the spring constant k=48 N/m, the amplitude (maximum displacement) A=5.5m, the angular frequency w=7.6 rad/s, and the equation for the position (displacement) x=A cos(wt).
(a) The angular frequency is related to the spring constant and the mass (m) of the particle by the equation
w=sqrt(k/m) => m=k/w^2=(48 N/m)/(7.6/s)^2=0.831 kg (Note: 1 N=1 kg m/s^2)
(b) The angular frequency is related to the period (T) by
w=2 pi/T => T=2 pi/w=2*3.14/(7.6/s)=0.826 s
(c) The instantaneous velocity (v) of an object is given by the time derivative of its position, i.e.,
v=dx/dt=d[A cos(wt)]/dt=A d[cos(wt)]/dt=A w [sin(wt)]=A w sin(wt)
The maximum speed occurs when sin(wt)=1, so that
vmax=A w=5.5m*7.6/s=41.8m/s
(d) The potential energy (U) stored in a spring is given by
U=(k x^2)/2=k [(A cos(wt))^2]/2=k A^2 cos^2(wt)/2
The potential energy is a maximum when cos(wt)=+/1, so
Umax=(k A^2)/2=48 N/m*((5.5m)^2)/2=726 J (Note: 1 J=1 N m= 1 kg m^2/s^2)
(e) The total energy (E) of the system is the sum of the potential energy (U) and the kinetic energy (K). Recall that K=(m v^2)/2 and m=k/w^2, so that
K=[m (A w sin(wt))^2]/2=(k/w^2)[A^2 w^2 sin^2(wt)]/2=k A^2 sin^2(wt)/2
In part (d) we found that
U=k A^2 cos^2(wt)/2
Thus,
E=K+U=k A^2 sin^2(wt)/2+k A^2 cos^2(wt)/2=[(k A^2)/2][sin^2(wt)+cos^2(wt)]
Recall that
sin^2 (B)+cos^2 (B)=1
So
E=(k A^2)/2=Umax=726 J