Exercise 14.23 A guitar suring vibrates at a frequency of 440 Hz. A point at its

Question

Exercise 14.23 A guitar suring vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 mm and a phase angle of zero Part A Write an equation for the position of the center of the sering as a funcrion of time Assume z (t) in meters and t in seconds SubmitRequest Answer Part B What is the maximum value of the magnitude of the velocity of the center of the sring Express your answer using two significant figures m/s SubmitR Part C What is the maximum value of the magnitude of the acceleration of the center of the string Express your answer using two significant figures. m/s SubmitRe Part D The derivative of the acceleration with respect to time is a quantity called the jerk Write an equation for the jerk of the center of the string a a function of time. Assume j(t) in m/s and t in seconds. j(t) m/s Submit Request Answer Part E Find the maximum value of the magnitude of he jerk Express your answer using two significant figures m/s SubmitRequest Answer Next

Explanation / Answer

a)

Angular frequency

W=2pif =2pi*440 =2764.6 rad/s

Therefore

X(t)=ACos(Wt)

X(t)=0.003Cos(2764.6t) m

b)

Maximum Velocity

Vmax=AW=0.003*2764.6 =8.294 m/s

c)

acceleration

amax=W2A =2764.62*0.003=2.293*104 m/s2

d)

X=0.003Cos(2764.6t)

V=dX/dt-0.003*2764.6Sin(2764.6t) =-8.293SIn(2764.6t)

a=dV/dt =-8.293*2764.6Cos(2764.6t)=-(2.293*104)Cos(2764.6t)

J=da/dt =-(2.293*104)*(-2764.6)Sin(2764.6t)

J(t)=(6.34*107)Sin(2764.6t) m3/s

e)

Jmax=6.34*107 m3/s

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