A guitar string vibrates at a frequency of 430 Hz. A point at its center moves i

Question

A guitar string vibrates at a frequency of 430 Hz. A point at its center moves in SHM with an amplitude of 4.3 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) = 4.3 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? velocity m/s acceleration m/s^2 (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/s^3

Explanation / Answer

x(t) is defined as x(t) = A cos (t +) A= 4.3 mm , t is our variable and

angular velocity is defined as =2f =2(430 hz) = 2700.4rad/sec

x(t) = 4.3 mm cos ((2700.4rad/sec)t)

Remember mm is 10-3 m

Vmax = A = (.0043m) (2700.4 rad/sec) = 11.61m/s

Amax = A2 =(.0043m)(2700.4rad/sec)2 = 3.13x 104m/s2

Jerk = A3 = (.0043m)(2700.4rad/sec)3 = 8.46x 107 m/s3

For the equation take the decrate of x(t) three times, thisgives you

= -A3cos(t + )

= -4.3mm(2700.4 rad/sec)3 cos ((2700.4rad/sec)t)

Try taking the derivatives yourself too ! Use chain rule andproduct rule.

Hope this helped!

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