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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(4.00Ï€x-400Ï€t)y2(x,t)=(6.00mm)sin(4.00Ï€³æ+400Ï€³Ù)

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?

Short Answer

Expert verified

The distance through which each wave moves along the string in the time interval that the point takes to move from the maximum upward displacement to maximum downward displacement is

Step by step solution

01

Given data

Two waves that create a standing wave are given:

y1(x,t)=(6.00mm)sin(4.00Ï€³æ-400Ï€³Ù)y2(x,t)=(6.00mm)sin(4.00Ï€³æ+400Ï€³Ù)

02

Understanding the concept of displacement of the wave

We can find the time taken by the wave to move from maximum upward displacement to maximum downward displacement in terms of time period. Then, using the relation between T and angular speed we can write an expression for time in terms of angular speed. Also, we can write the velocity in terms of angular speed. Then, from the velocity and time, we can easily calculate the distance travelled by the wave along the string in the time interval that the point takes to move from maximum upward displacement to maximum downward displacement.

Formulae:

The time period of oscillation, T=2Ï€Ó¬...........(1)

The velocity of the wave, v=Ó¬k...........(2)

The displacement change of the wave, ∆x=vt......(3)

03

Calculation of the maximum downward displacement

Two waves that create the standing wave are

y1(x,t)=(6.00mm)sin(4.00Ï€³æ-400Ï€³Ù)y2(x,t)=(6.00mm)sin(4.00Ï€³æ+400Ï€³Ù)

Therefore, according to the superposition principle, the equation of the resultant wave is

y'=2ymsinkxcosÓ¬t=12mmsin4.00Ï€³æcos400Ï€³Ù.........(4)

The time taken by the wave to move from maximum upward displacement to maximum downward displacement is T/2.

t=T2=2Ï€2Ó¬fromequation(1)=Ï€Ó¬

Substituting the value of time and velocity from equation (2) in equation (3), we get the displacement as:

∆x=ӬkπӬ=πk=π4.00π(fromequation(4),wegetk=4.00π)

Therefore, the distance through which each wave moves along the string in the time interval that the point takes to move from maximum upward displacement to maximum downward displacement is 0.25 m

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Most popular questions from this chapter

Figure 16-25agives a snapshot of a wave traveling in the direction of positive xalong a string under tension. Four string elements are indicated by the lettered points. For each of those elements, determine whether, at the instant of the snapshot, the element is moving upward or downward or is momentarily at rest. (Hint:Imagine the wave as it moves through the four string elements, as if you were watching a video of the wave as it traveled rightward.) Figure 16-25bgives the displacement of a string element located at, say, x=0as a function of time. At the lettered times, is the element moving upward or downward or is it momentarily at rest?

These two waves travel along the same string:

y1(x,t)=(4.60mm)sin(2Ï€³æ-400Ï€³Ù)y2(x,t)=(5.60mm)sin(2Ï€³æ-400Ï€³Ù+0.80Ï€°ù²¹»å)

What is the amplitude (a) and (b) what is the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 5.00 mmis also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?

In Figure 16-36 (a), string 1 has a linear density of 3.00 g/m, and string 2 has a linear density of 5.00 g/m. They are under tension due to the hanging block of mass M = 500 g. (a)Calculate the wave speed on string 1 and (b) Calculate the wave speed on string 2. (Hint:When a string loops halfway around a pulley, it pulls on the pulley with a net force that is twice the tension in the string.) Next the block is divided into two blocks (with M1+M2=M) and the apparatus is rearranged as shown in Figure (b). (c) Find M1and (d) Find M2such that the wave speeds in the two strings are equal.

The functiony(x,t)=(15.0cm)cos(ττ³æ-15ττ³Ù), with x in meters and t in seconds, describes a wave on a taut string. What is the transverse speed for a point on the string at an instant when that point has the displacement y=+12.0cm?

Use the wave equation to find the speed of a wave given in terms of the general function: h(x,t)

localid="1660990709658" y(x,t)=(4.00mm)h[(30m-1)x+(6.0s-1)t].

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