Question

Please answer part d).

A string is tied between two posts sticking out of the wall. The posts are separated by 1.20 m. A 5.00-cm-long snippet of the string weighs 1.50 grams. You attach a machine to one of the posts, so that the post can vibrate up and down at a frequency set on the machine. (You can still consider the string attached to this post as a node - the machine does not create a high amplitude right at the post.)

a) You slowly turn up the frequency of the machine, and at 41.7 Hz you see the standing-wave pattern corresponding to the fundamental frequency of the system. What is the tension in the string? Express the answer in newtons (N). ANSWER: T = 300 N

b) You continue to increase the frequency of the machine, and at a higher frequency you observe that nodes form every 10 cm, and the antinodes have an oscillation amplitude of 5.2 mm. What is the frequency of this oscillation? Express the answer in hertz (Hz). ANSWER: f = 500 Hz

c) Another standing wave pattern appears when the machine is set to 167 Hz. The amplitude of the string's oscillation at the antinodes is 5.2 mm. How many antinodes n does the pattern have? ANSWER: n = 4.00

d) You keep the conditions set as they are in part C. What is the amplitude of the string's oscillation 40 cm from the post that the machine is oscillating? Express the answer in millimetres (mm).

this q has been posted before and all answers correct with exception to d

Answer 1

d)keeping the same conditions,We have the standing wave formed in the string of 4th harmonic,ie,

so we have the amplitude of the standing wave,given as,(froms at antinodes)

Also we have the wavw eqaution of standin waves which is formed by the superimposition of the 2 identical waves of same parameters such as frequency,amplitude,wavwlength...moving in opposite direction to each other.

So,We have the wave equation of standing wave,

So,Here,we have At 4th harmonic, So,The string contains the standing wave with 5 nodes and 4 anitinodes as shown in figure.(ie,4th harmonic,n=4)

So,there are 2 wavelength of the wave in the length of the string between 2 posts.

ie,

length of string,L=1.2m

Or wave length of standing wave in this harmonic,

Consider the time initially as initially to solve as the standing waves are stationary and doesnot moves with time or change its position.

So here quation beocmes, (since cos0=1)

So,for finding the displacement() of the standing wave at the distance x=40cm from the post,

We have,

So,Eqn ,

So,the vertical displacement from x axis at x=40cm from post,

Or the amplitude or the vertical displacement of the wave(string osscillation) at x=40cm from the post,4.5cm