Q9P Question: If the form of a sound... [FREE SOLUTION]

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Chapter 17: Q9P (page 506)

Question: If the form of a sound wave traveling through air is $s (x, t) = (6.0 n m) \ c o s (kx + (3000 \frac{r a d}{s}) t + f)$ .How much time does any given air molecule along the path take to move between displacement s = + 2.0 nmand s = - 2.0 nm?

Short Answer

Expert verified

Answer

The time taken by the air molecule to move between displacements to s = + 2.0 nmand s = - 2.0 nmis 0.23 ms .

Step by step solution

Given data

The form of sound wave, $s (x, t) = (6.0 nm) \cos (k x + 3000 \frac{rad}{s} t + 蠁)$ .
The molecule move between s = + 2.0 nmand s = - 2.0 nm.

Determining the concept

By using the values of displacements in the given form of the sound wave, two equations can be found. By subtracting and solving them, find the time taken by the air molecule to move between displacements s = + 2.0 nmand s = - 2.0 nm..

Determining the time does any given air molecule along the path take to move between displacement is s = + 2.0 nm and s = - 2.0 nm.

$1.90 = k x + 3000 \frac{rad}{s} t_{2} + 蠁$ The form of the sound wave, $s (x, t) = (6.0 nm) \cos (k x + 3000 \frac{rad}{s} t + 蠁)$ .

At $s = + 2.0 nm,$

$+ 2.0 nm = (6.0 nm) \cos (k x + 3000 \frac{rad}{s} t_{1} + 蠁) \frac{2.0nm}{6.0nm} = \cos (k x + 3000 \frac{rad}{s} t_{1} + 蠁) (\frac{2.0}{6.0}) = k x + 3000 \frac{rad}{s} t_{1} + 蠁 (\frac{1}{3}) = k x + 3000 \frac{rad}{s} t_{1} + 蠁 (0.33) = k x + 3000 \frac{rad}{s} t_{1} + 蠁$

$1.23 = k x + 3000 \frac{rad}{s} t_{1} + 蠁$ 鈥︹€�. (i)

At, $s = - 2.0 nm,$

role="math" localid="1661754951618" $- 2.0 nm = (6.0 nm) \cos (k x + 3000 \frac{rad}{s} t_{2} + 蠁) \frac{- 2.0nm}{6.0nm} = \cos (k x + 3000 \frac{rad}{s} t_{2} + 蠁) (\frac{- 2.0}{6.0}) = k x + 3000 \frac{rad}{s} t_{2} + 蠁 (- \frac{1}{3}) = k x + 3000 \frac{rad}{s} t_{2} + 蠁 (- 0.33) = k x + 3000 \frac{rad}{s} t_{2} + 蠁$

$1.90 = k x + 3000 \frac{rad}{s} t_{2} + 蠁$ 鈥︹€�. (ii)

Subtracting equation 1 from 2,

$1.90 - 1.23 = 3000 \frac{rad}{s} t_{2} - 3000 \frac{rad}{s} t_{1} 0.67 = 3000 (t_{2} - t_{1}) t_{2} - t_{1} = \frac{0.67}{3000} t_{2} - t_{1} = 0.00022$

$鈭� t_{2} - t_{1} = 0.23 ms$

Therefore the time taken by the air molecule to move between displacements to s = + 2.0 nm and s = - 2.0 nm is 0.23 ms .

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91影视

Question: If the form of a sound wave traveling through air is $s (x, t) = (6.0 n m) \ c o s (kx + (3000 \frac{r a d}{s}) t + f)$ .How much time does any given air molecule along the path take to move between displacement s = + 2.0 nmand s = - 2.0 nm?

Short Answer

Step by step solution

Given data

Determining the concept

Determining the time does any given air molecule along the path take to move between displacement is s = + 2.0 nm and s = - 2.0 nm.

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91影视

Question: If the form of a sound wave traveling through air iss(x,t)=(6.0nm)\cos(kx+3000radst+f).How much time does any given air molecule along the path take to move between displacement s = + 2.0 nmand s = - 2.0 nm?

Short Answer

Step by step solution

Given data

Determining the concept

Determining the time does any given air molecule along the path take to move between displacement is s = + 2.0 nm and s = - 2.0 nm.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Classical Mechanics

Translational Dynamics

Physics of Motion

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Rotational Dynamics

Study anywhere. Anytime. Across all devices.

Question: If the form of a sound wave traveling through air is $s (x, t) = (6.0 n m) \ c o s (kx + (3000 \frac{r a d}{s}) t + f)$ .How much time does any given air molecule along the path take to move between displacement s = + 2.0 nmand s = - 2.0 nm?