Q13E A transverse wave on a string ha... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 2: Q13E (page 498)

A transverse wave on a string has amplitude 0.300 cm, wavelength 12.0 cm, and speed 6.00 cm/s. It is represented by y(x,t) as given in Exercise 15.12.
(a) At time t = 0, compute y at 1.5-cm intervals of x(that is, at x = 0, x = 1.5 cm, x = 3.0 cm, and so on) from x= 0 to x = 12.0 cm. Graph the results. This is the shape of the string at time t = 0.
(b) Repeat the calculations for the same values of x at times t = 0.400 s andt = 0.800 s. Graph the shape of the string at these instants. In what direction is the wave traveling?

Short Answer

Expert verified

(a)(i)

x (cm)	y (cm)
0	0.3
1.5	0.212
3	0
4.5	-0.212
6	-.0.3
7.5	-0.212
9	0
10.5	0.212
12	0.3

(ii)The graph is drawn below

Step by step solution

Given Data

\(\begin{array}{l}{\rm{Amplitude}},\;A=0.3\;{\rm{cm}}\\{\rm{wavelength}},\,\;\lambda=12\;{\rm{cm}}\\{\rm{speed}},\;v= 6\;{\rm{cm/s}}\end{array}\)

Concept

The motion of a transverse wave is occurred, when all the points in the wave oscillate in the direction of right angle to the path of wave,

Step 3(a)(i): compute y

\(\begin{array}{c}x = 0\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( 0 \right)}}} \right.\\} {12 \times \left( 0 \right)}}} \right)\\ = 0.3\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 1.5\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( {1.5} \right)}}} \right.\\} {12 \times \left( {1.5} \right)}}} \right)\\ = 0.212\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 3\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( 3 \right)}}} \right.\\} {12 \times \left( 3 \right)}}} \right)\\ = 0\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 4.5\;{\rm{cm}}\\x = 1.5\;{\rm{cm}}\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( {4.5} \right)}}} \right.\\} {12 \times \left( {4.5} \right)}}} \right)\\ = 0.212\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 6\;{\rm{cm}}\\x = 1.5\;{\rm{cm}}\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( 6 \right)}}} \right.\\} {12 \times \left( 6 \right)}}} \right)\\ = - 0.3\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 7.5\;{\rm{cm}}\\x = 1.5\;{\rm{cm}}\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( {7.5} \right)}}} \right.\\} {12 \times \left( {7.5} \right)}}} \right)\\ = - 0.212\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 9\;{\rm{cm}}\\x = 1.5\;{\rm{cm}}\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( 9 \right)}}} \right.\\} {12 \times \left( 9 \right)}}} \right)\\ = 0\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 10.5\;{\rm{cm}}\\x = 1.5\;{\rm{cm}}\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( {10.5} \right)}}} \right.\\} {12 \times \left( {10.5} \right)}}} \right)\\ = 0.212\;{\rm{cm}}\end{array}\)\(\begin{array}{c}x = 12\;{\rm{cm}}\\x = 1.5\;{\rm{cm}}\\y\left( {x,t} \right) = A\cos \;\left( {\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } \lambda }} \right.\\} \lambda }} \right)\left( {x - vt} \right)} \right)\\y = 0.3 \times \cos \;\left( {{{2\pi } \mathord{\left/{\vphantom {{2\pi } {12 \times \left( {12} \right)}}} \right.\\} {12 \times \left( {12} \right)}}} \right)\\ = 0.3\;{\rm{cm}}\end{array}\)

Hence the values are