Question

At t = 0, the instantaneous position of two pulses moving along a taut string with a speed v = 23.2 cm/s are as shown in the diagram below. Each unit on the horizontal axis is 4.0 cm and each unit on the vertical axis is 3.0 cm. (The peak of pulse 2 is exactly on a half unit of the horizontal axis.)

(a) At what location will the resultant of the two pulses have maximum amplitude?
cm

(b) At what time will the resultant of the two pulses have maximum amplitude?
s

(c) What is the value of this maximum amplitude?
cm

2. A guitar string of length L is bound at both ends. The table shows the string's harmonic frequencies when struck.

Harmonic Number	Frequency
1	217 L
2	434 L
3	651 L
4	868 L

(a) Based on the information above, what is the speed of the wave (in m/s) within the string?

m/s

(b) The guitarist then slides her finger along the neck of the guitar, changing the string length as a result. Calculate the fundamental frequency of the string if the string length is reduced to

5/6 L

f1=

Calculate the wave speed present (in m/s) if the string length is reduced to 5/6 L

m/s L.

Answer 1

According to HOMEWORKLIB POLICY we have to solve the first question

Solution 9.5X y The distance blue the there fulses 2 19cm (al at IC= 2x 4 + 19 8 + 19 = 27cm (6) tad 19 li = 0.8188 v 23:2 Resultare amplitude A : 2X3+ 3x3 - 15 cm