A 0.173-kg particle undergoes simple harmonic motion along the horizontal x-axis

Question

A 0.173-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.347 m and x2 = 0.391 m. The period of oscillation is 0.555 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy,

Etot. f= ______ Hz

Xeq= _____ m

A= ________ m

Vmax= ________ m/s

Amax=_________ m/s^2

k=________ N/m

Etotal=_______ J

This is the first response from an anonymous person on chegg, and it is partially incorrect.

f=1/T=1/0.555=1.801 s

xeq=(x1+x2)/2=(-0.347+0.391)/2=0.022 m

A=(x2-x1)/2=(0.391+0.347)/2=0.369 m

x(t)=Acos(wt+) v(t)=-wAsen(wt+) w=2f=2*3.14*1.801=11.31rad/s

vmax=-wA=11.31*0.369=-4.17m/s

a(t)=-w^2Acos(wt+) amax=-w^2A=96.04*0.369=-35.43 m/s^2

w=(k/m)^1/2 => k=mw^2=0.173*96.04=16.61 N/m

Etot=1/2kA^2 Etot=0.5*16.61*0.136=1.12 J

amax, Etot, and k are incorrect.

Explanation / Answer

(a)
f = 1/T
f = 1/0.555
f = 1.801 Hz

(b)
xeq = (x1+x2)/2
xeq = (-0.347+0.391)/2
xeq = 0.022 m

(c)
A = (x2-x1)/2
A = (0.391+0.347)/2
A = 0.369 m

(d)
x(t) = Acos(wt+)
Where,
w = 2f
w = 2*3.14*1.801
w =11.31 rad/s

v(t)= - wA sin(wt+)
vmax = -wA
vmax = - 11.31*0.369
vmax = - 4.17 m/s

(e)
a(t) = -w^2A cos(wt+)
amax = -w^2*A
amax = - 11.31^2 * 0.369
amax = - 47.2 m/s^2

(f)
w= sqrt(k/m)
11.13 = sqrt(k/0.173)
k = 21.43 N/m

(g)
Etot = 1/2 *k * A^2
Etot = 1/2 *21.43 *0.369^2
Etot =1.46 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.