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The following four waves are sent along strings with the same linear densities (xis in meters and tis in seconds). Rank the waves according to (a) their wave speed and (b) the tension in the strings along which they travel, greatest first:

(1)Y1=(3mm)sin(x-3t), (3)y3=(1mm)sin(4x-t),

(2) y2=(6mm)sin(2x-t), (4)y4=(2mm)sin(x-2t).

Short Answer

Expert verified
  1. The waves can be ranked according to their wave speed as V1>V4>V2>V3(greatest first)
  2. The waves can be ranked according to their tension in the string along which they travel asT1>T4>T2>T3 (greatest first)

Step by step solution

01

Step 1: Given

The four waves along the strings with the same linear densities are,

Y1=3mm)sin(x-3t)y2=(6mm)sin(2x-t)y3=(1mm)sin(4x-t)y4=(2mm)sin(x-2t)

02

Determining the concept

Use the concept of the equation of transverse wave and speed of a travelling wave. The wave speed on a stretched string gives the relation between speed and tension in the string.

Formulae are as follow:

y=ymsin(kx-ฯ‰t)v=ฯ‰kv=Tฮผ

Where, v is wave speed, T is tension in string, ๐ is mass per unit length, ๐Ž is angular frequency, k is wave number, t is time.

03

Determining to rank the waves according to their wave speed

(a)

Rank the waves according to their wave speed :

The equation of transverse wave is,

y=ymsin(kx-ฯ‰t)...........................................................................(1)

The speed of the travelling wave is,

v=wk

The equation (i) is,

y1=(3mm)sin(x-3t)

Compare this equation with equation (1), then the speed of the travelling wave is,

v1=31v1=3

The equation (ii) is,

y2=(6mm)sin(2x-t)

Compare this equation with equation (1), then the speed of the travelling wave is,

V2=12v2=0.5

The equation (iii) is,

y3=(1mm)sin94x-t)

Compare this equation with equation (1), then the speed of the travelling wave is,

V3=14V3=0.25

The equation (iv) is,

y4=(2mm)sin(x-2t)

Compare this equation with equation (1), then the speed of the travelling wave is

V4=21V4=2

Hence, the rank of the waves according to the wave speed is V1>V4>V2>V3 (greatest first).

04

Determining to rank the waves according to the tension in the string along which they travel

(b)

Rank the waves according to tension:

The wave speed on a stretched string is,

V=TฮผVโˆT

The speed on the stretched string is directly proportional to the tension in the string with the same linear density.

The speed on the stretched string for equation (i) is,

v1โˆT1

The speed on the stretched string for equation (ii) is,

V2โˆT2

The speed on the stretched string for equation (iii) is,

V3โˆT3

The speed on the stretched string for equation (i) is,

V4โˆT4

Hence, the rank of the waves according to their tension is T1>T4>T2>T3 (greatest first).

Therefore, the wave speed by using its expression and rank their values can be found, by using the expression of the speed on the stretched string, and can also find the tension in each string and rank their values.

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Most popular questions from this chapter

A 120 mlength of string is stretched between fixed supports. What are the (a) longest, (b) second longest, and (c) third longest wavelength for waves traveling on the string if standing waves are to be set up? (d)Sketch those standing waves.

Four waves are to be sent along the same string, in the same direction:

y1(x,t)=(4.00mm)sin(2ฯ€x-400ฯ€t)y2(x,t)=(4.00mm)sin(2ฯ€x-400ฯ€t+0.7ฯ€)y3(x,t)=(4.00mm)sin(2ฯ€x-400ฯ€t+ฯ€)y4(x,t)=(4.00mm)sin(2ฯ€x-400ฯ€t+1.7ฯ€)


What is the amplitude of the resultant wave?

Figure 16-32 shows the transverse velocity u versus time t of the point on a string at x = 0 , as a wave passes through it. The scale on the vertical axis is set by us=4.0m/s . The wave has the form y(x,t)=ymsin(kx-ฯ‰t+ฯ•) . What then is ฯ• ? (Caution:A calculator does not always give the proper inverse trig function, so check your answer by substituting it and an assumed value of ฯ‰ into y(x,t)and then plotting the function.)

The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane:

y1(x,t)=(6.00mm)sin(4.00ฯ€x-400ฯ€t)y2(x,t)=(6.00mm)sin(4.00ฯ€x+400ฯ€t)

within X meters andin seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

A string oscillates according to the equationy'=(0.50cm)sin[(ฯ€3cm-1)x]cos[(40ฯ€s-1)t]What are the (a) amplitude and (b) speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation? (c) What is the distance between nodes? (d) What is the transverse speed of a particle of the string at the positionx=1.5cmwhent=98s?

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