How far, and in what direction, should a cellist move her finger to adjust a string's tone from an out-of-tune 453 Hz to an in-tune 440 Hz? The string is 68.0 cm long, and the finger is 18.3 cm from the nut for the 453-Hz tone. (Assume the tone comes from the portion of the string between the cellist's finger and the end opposite the nut. Assume the string vibrates in the fundamental mode. Take the direction towards the nut to be positive. Indicate the direction with the sign of your answer.)

She should move her finger 14 cm towards the nut

Explanation:

First of all let's find the velocity with the given wavelength and initial frequency (the outtune one ).

The formula is;

Velocity = Wavelength x frequency

We are given;

Since the string is 68.0 cm long, and the finger is 18.3 cm from the nut . Thus wavelength = 68 - 18.3 = 49.7cm = 0.497m

The outtune frequency is 453 Hz

Thus;

velocity = 0.497*452

Velocity = 224.644 m/s

We will now use this velocity in the same equation, but substitute the frequency value for the intune one and then use the new wavelength to tell where the cellist should move her finger

Thus;

velocity = wavelength*frequency

Wavelength = velocity/frequency

Thus;

wavelength = 224.644/440

Wavelength = 0.511 m

Difference between the 2 wavelengths is;

0.511m - 0.497m = 0.014m or 14cm (distance)

Since the string is actually increasing, she should move her finger 14 cm towards the nut

think about the oscillation of the toy shown in the video. when the toy is at the highest point of its motion, its velocity is zero and its acceleration is negative.