Question

Two ropes with different linear densities but the same length of L -1.58 m are joined together. The linear density of the first rope is 0.05670 kg/m. A force of F 1620 N is supplied to the ropes to keep them taut Figure 1 The frequency of a sinusoidal frequency generator is adjusted until two loops can be seen in the first rope and three loops in the second rope 2 Sinusoidal frequency generator First rope Second rope Figure 1: The two different ropes joined together at one end, with the other end of the first rope connected to the sinusoidal frequency generator. A force, F, is applied to the other end of the second rope, directed away from the sinusoidal frequency generator Part 1) What is the frequency of the sinusoidal frequency generator when this pattern is observed? f107 Hz Your last answer was interpreted as follows: 107 Part 2) What is the linear density of the second rope? kg/m Your last answer was interpreted as follows: 0.128 Part 3) The force, F, applied on the end of the two ropes is now gradually reduced. What is the magnitude of the force applied the next time standing waves are simultaneously observed in the two ropes?

I couldn't figure out part 3 of this question.

Answer 1

Part 1)For the first rope, two waves formed, $L=n\frac{\lambda }{2}=2\frac{\lambda}{2}$

1620_ = 107 Hz バー臺副--2+258v0 052070 =107 H: ひ ひ 2*1.58 V 0.05670 2L μι 2L

Part 2) For the second rope, three waves are formed, $L=n\frac{\lambda }{2}=3\frac{\lambda}{2}$

$f=\frac{v}{\lambda}=3\frac{v}{2L}=\frac{3}{2L}\sqrt{\frac{F}{\mu_2}}$

9F 16200.1275kg/m 4 * 1.582*1072 = 0.1275 kg/m

Part 3)

Now F is decreased.

Frequency of two strings remains the same,

For the first string, $f=\frac{n_1}{2L}\sqrt{\frac{F}{\mu_1}}$ ---(1)

For the second string $f=\frac{n_2}{2L}\sqrt{\frac{F}{\mu_2}}$ ---(2)

Dividing above two equations,  

$\frac{\frac{n_2}{2L}\sqrt{\frac{F}{\mu_2}}}{\frac{n_2}{2L}\sqrt{\frac{F}{\mu_2}}}=1$

$\frac{n_1}{n_2}=\sqrt{\frac{\mu_1}{\mu_2}}=\sqrt{\frac{0.05670}{0.1275}}=0.6666=\frac{2}{3}=\frac{4}{6}$

So the next time the standing waves are formed, $n_1=4\,n_2=6$

Using equation (1) $f=\frac{n_1}{2L}\sqrt{\frac{F}{\mu_1}} \Rightarrow F=\frac{4L^2f^2\mu_1}{n_1^2}$

$F=\frac{4(1.58)^2(107)^2(0.05670)}{4^2}=405\,N$