Prof. Williams
2.3.2020
Standing Waves on a String
V. CALCULATIONS
1. Calculate the density of the string.
m 0.00226 =n = 0.001609 kg/m 1.405
2. Calculate the wavelength for each standing wave situation.
2 * 1.083 1 = = 2.166 m 1
3. Compute the speed of the wave using the tension and mass density.
T
mg 6 u
0.5 * 9.81 55.2209 m/s 0.001609
u
4. Compute the speed of the wave by using the wavelength and frequency.
v = lf = 2.166 * 26.2 = 56.7492 m/s
5. Compare the values computed in steps 3 and 4.
In step 3, all the speeds are 55.2209 m/s
In step 4 the speed are roughly close to each other and their average is 56.8671 m/s
56.8671-55.2209 * 100 = 2.93%
to the computed values in step 4.
VI. CONCLUSION/QUESTIONS
1. What are the sources of error in the experiment?
-There is a possibility of not using the exact best frequency to get desired nodes
-Error in measuring the distances between nodes
-The string may not be set up in a perfect straight line
2. Is there an actual node in the string at the vibrator? Explain.
There is not an actual node in the string because by the definition there is no displacement at
however, at the vibrator position the particles are constantly moving and vibrating
3. What is the purpose of wrapped strings used in some of musical instruments?
In a musical instrument we would like to play different frequencies hence we must have different
T wave speeds. We know that v = - so we either have to change mass density or tension. VA
Consequently, the string would be very loose for low frequencies and very tight for high
frequencies, and it is not convenient to play in this case. It would be more convenient if we can
play with the strings that roughly have the same tension. Based on the formula the lower strings
must have higher mass density, so we need to make the lower ones thicker. For this purpose, we
can wrap the low strings them with wire.
4. What conclusions can you reach about the relationship between the speed of the wave on
the string and the number of antinodes in the standing wave pattern?
In the case that we have fixed mass, as number of antinodes increase, the speed of the wave
remains the same because v = I, and since we are dealing with fixed mass speed remains
constant. Consequently, as number of antinodes increase, the speed of the waves remains
constant. We know that 2L/n = 1, so increasing the number of antinodes will decrease the wavelength.
however the frequencies get larger and because v = f, the speed of the wave roughly remains constant.
The calculated speed using v = Af are not exactly qual to the values using v =
because of
In the case with fixed frequency, increasing the number of the antinodes will decrease the
wavelengths (2L