Problem 4 String 1 in the figure has linear density 2.70 g/m and string 2 has li

Question

Problem 4 String 1 in the figure has linear density 2.70 g/m and string 2 has linear density 3.20 g/m. A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. She wants both pulses to reach the ends of the strings simultaneously (Figure 1) Part A What should the string length L1 be? Express your answer with the appropriate units. Value Submit Part B What should the string length L2 be? Express your answer with the appropriate units. alue Figure 1 of 1 Submit String 1 String 2 Return to Assignment Provide Feedback Knot

Explanation / Answer

  the key piece of physics here is that the speed of a wave on a string is

v=sqrt[T/u] where T is the tension in the string and u is the mass/length of the string

the set up of the experiment makes it sound as if the tension is the same in both strings, so using subscripts 1 and 2 to refer to properties of string 1 and 2, we have

v1=sqrt[T/u1]
v2=sqrt[T/u2]

which gives the ratio

v1/v2=sqrt[u2/u1]=sqrt[3.2/2.7]=1.088

v1=1.088v2

now, if the pulses reach the end at the same time, we know from the simple distance = speed x time that

time = dist/speed so that

t1=L1/v1 and t1=L2/v2

but since t1=t2, L1/v1=L2/v2

or L1 v2=L2 v1

we know that v1=1.088v2, so

L1 v2 = L2(1.088 v2) or L1=1.088L2

but since L1+L2=4, this tells us

1.088L2 + L2=4 or L2=1.915 and L1=2.08

Hence Part A L1 = 2.08 m

and Part B L2 = 1.915 m

Hope this helps you. Please rate the answer.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.