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Chapter 39: Q.18 (page 1137)

akjsbdl

Short Answer

Expert verified

So the length of the pulse will be $85 m$

Step by step solution

Given information

Sound waves of 498 Hz and 502 Hz are superimposed at a temperature where the speed of sound in air is 340 m/s. What is the length ∆x of one wave packet?

Step:2 Calculation

Here $Δ f = 502 - 498 = 4 Hz$

So $Δ t = \frac{1}{4} = 0.35 s .$

So $Δ x = 0.25 \times 340 = 85 m$ .

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Most popular questions from this chapter

FIGURE P39.31 shows the wave function of a particle confined

between x = 0 nm and x = 1.0 nm. The wave function is zero

outside this region.

a. Determine the value of the constant c, as defined in the figure.

b. Draw a graph of the probability density $P (x) = {|ψ (x)|}^{2}$

c. Draw a dot picture showing where the first 40 or 50 particles

might be found.

d. Calculate the probability of finding the particle in the interval

$0 n m \leq x \leq 0.25 n m$ .

What are the units of $ψ (x)$ ? Explain.

A 1.0-mm-diameter sphere bounces back and forth between two walls at $x = 0 mm$ and $x = 100 mm$ . The collisions are perfectly elastic, and the sphere repeats this motion over and over with no loss of speed. At a random instant of time, what is the probability that the center of the sphere is

a. At exactly $x = 50.0 mm$ ?

b. Between $x = 49.0 mm$ and $x = 51.0 mm$ ?

c. At $x \geq 75 mm$ ?

The wave function of a particle is

$ψ (x) = \sqrt{\frac{b}{π (x^{2} + b^{2})}}$

where b is a positive constant. Find the probability that the particle is located in the interval -b $\leq$ x $\leq$ b

The probability density for finding a particle at position $x$ is

$p (x) = \{\begin{cases} \frac{a}{(1 - x)} \\ b (1 - x) \end{cases} \frac{- 1 m m \leq x < 0 m m}{0 m m \leq x \leq 1 m m}$

and zero elsewhere

See all solutions

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akjsbdl

Short Answer

Step by step solution

Given information

Step:2 Calculation

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Most popular questions from this chapter

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