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Chapter 17: Q. 39 (page 484)

Biologists think that some spiders “tune” strands of their web to give an enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20 µm, and spider silk has a density of 1300 kg/m3. To have a fundamental frequency at 100 Hz, to what tension must a spider adjust a 12-cm-long strand of silk?

Short Answer

Expert verified

The tension in the silk strand is 2.35 x 10-4 N

Step by step solution

01

Write the given information

Diameter of spider web d= 20 µm =20 x 10-6 m
Radius of the spider web, r= 10-5m
The density of the spider silk, ρ= 1300 kg/ m3
The frequency of the silk strand, f = 100Hz
Length of the silk strand, L= 12 cm= 0.12m

02

 Step 2: To determine the tension in the silk

The mass per unit length of the silk is given by
µ=ρAµ=ρ(πr2)=13003.1410-52=4.082x10-7kg/m

The fundamental frequency of the wave is given by
localid="1650099139558" f=12LTµ


Substitute the known values

localid="1650099446065" f=12LTµ100=12(0.12)T4.082x10-7kg/m24=T4.082x10-7

Squaring on both the sides
576=T4.082x10-7T=2.35x10-4N

Thus the tension in the silk strand is 2.35 x 10-4 N

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Most popular questions from this chapter

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x = 0.50 m. If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at x = 0.90 m.

a. What is the frequency of the sound? Assume vsound = 340 m/s.

b. What is the phase difference between the speakers?

Two loudspeakers in a plane, 5.0 m apart, are playing the same frequency. If you stand 12.0 m in front of the plane of the speakers, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 12.0 m in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers. What is the frequency of the sound? Assume a sound speed of 340 m/s.

||| A water wave is called a deep-water wave if the water’s depth

is more than one-quarter of the wavelength. Unlike the waves

we’ve considered in this chapter, the speed of a deep-water wave

depends on its wavelength:

v = B

gl

2p

Longer wavelengths travel faster. Let’s apply this to standing waves.

Consider a diving pool that is 5.0 m deep and 10.0 m wide. Standing

water waves can set up across the width of the pool. Because

water sloshes up and down at the sides of the pool, the boundary

conditions require antinodes at x = 0 and x = L. Thus a standing

water wave resembles a standing sound wave in an open-open tube.

a. What are the wavelengths of the first three standing-wave

modes for water in the pool? Do they satisfy the condition for

being deep-water waves?

b. What are the wave speeds for each of these waves?

c. Derive a general expression for the frequencies fm of the possible

standing waves. Your expression should be in terms of m, g, and L.

d. What are the oscillation periods of the first three standing wave

What are the three longest wavelengths for standing sound waves in a 121-cm-long tube that is (a) open at both ends and (b) open at one end, closed at the other?

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