Standing wave velocity/acceleration

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A string vibrating in its third harmonic has a wave speed of 195 m/s and a frequency of 240 Hz, with an antinode amplitude of 0.350 cm. The amplitude at a point 24 cm from the left was calculated to be 0.00336 m, and the time for the string's motion from maximum upward to maximum downward displacement is 0.00208 s. To find the maximum transverse velocity and acceleration, the equation 2Asin(kx)sin(wt) is used, where ω can be derived from the time calculation. The maximum values occur when the sinusoidal function reaches its peak, and derivatives with respect to time yield the necessary velocity and acceleration. Understanding the position-dependent amplitude and time's role in the equation is crucial for accurate calculations.
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A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 195 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.350 cm.

I have already correctly calculated the amplitude of the string at a point 24 cm from the left of the string to be .00336 m and time it takes the string to go from its largest upward displacement to its largest downward displacement at that point to be .00208 s.

I need to calculate the maximum transverse velocity and acceleration of the string at that point. I know the equation I need to use is 2Asin(kx)sin(wt)

(((w is omega)))

I know how to take the partial derivates of the equation too, but I don't know what to do about the t in the equation. My textbook seems to ignore it, and there's nothing about the time in the equation. The textbook also ignores the x, but I am given a value of x to be 24 cm. I'm also down to my last guess, so I want to be cautious.
 
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Each point on he string moves as a harmonic oscillator with the same frequency. The amplitude is position dependent In your equation

2Asin(kx)sin(wt)

2A is the amplitude at an antinode (A being the amplitude of the waves traveling in opposite directions) and 2Asin(kx) is the amplitude of the motion at location x. If your time calculation is correct you have the time for one half-period of motion, so you can calculate ω. To get the velocity take one derivative with respect to time (2Asin(kx) is a constant for this derivative) and to get the acceleration take the derivative of the velocity. Each derivative adds a factor of ω in front of the resulting sin(ωt) or cos(ωt) times the position dependent amplitude that is already there. The maximum velocity and acceleration occur when the sinusoidal time function has its maximum value of one.
 
I have nearly the same question and had no issues calculating the time from top to bottom, but i wondered if you could explain how you found the amplitude at a given x without a given time.
 
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