A guitar string vibrates at a frequency of 380 Hz. A point at its center moves i

Q: A guitar string vibrates at a frequency of 380 Hz. A point at its center moves i

a) the displacement has to be zero at t=0, so you need to use the trig function that returns 0 when t=0, so try x(t) = 0.0029 sin(w t) b) v = 2 * pie * f * A v = 2 * 3.14 * 380 * 0.0029 v = 6.92 m/s then the acceleration is a = w^2 * A a = ( 2 * 3.14 * 380 )^2 * 0.0029 a = 1.65 * 10^4 m/s^2

ID: 1352886 • Letter: A

Question

A guitar string vibrates at a frequency of 380 Hz. A point at its center moves in SHM with an amplitude of 2.9 mm and a phase angle of zero.

(a) Write an equation for the position of the center of the string as a function of time t. (Assume x(0) = 2.9 mm. Keep the value of the amplitude in mm. Use the following as necessary: t.)
x(t) =

(b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string?

(c) 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 as a function of time. (Keep the value of the amplitude in mm. Use the following as necessary: t.)
j(t) =

Find the maximum value of the magnitude of the jerk.
___m/s3

velocity ____ m/s acceleration ___ m/s2

Explanation / Answer

the displacement has to be zero at t=0, so you need to use the trig function that returns 0 when t=0, so try x(t) = 0.0029 sin(w t)

v = 2 * pie * f * A

v = 2 * 3.14 * 380 * 0.0029

v = 6.92 m/s

then the acceleration is

a = w^2 * A

a = ( 2 * 3.14 * 380 )^2 * 0.0029

a = 1.65 * 10^4 m/s^2

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A guitar string vibrates at a frequency of 380 Hz. A point at its center moves i

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