Home Page Textbooks Physics: Principles and Problems Page 401: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 401: Section Review

Exercise 1

Step 1

1 of 2

Information given in the text are:

$E_{p} = 8.67, mathrm{J}$

/$x = 247, mathrm{mm} = 0.247, mathrm{m}$

Energy is given by:

$$
E_{p} = dfrac{1}{2}kx^{2}
$$

We can express spring constant as:

$$
k= dfrac{E}{1/2x^{2}}
$$

When we put known values into the previous equation we get:

$$
k = dfrac{8.67, mathrm{J}}{1/2 cdot (0.247, mathrm{m})^{2}}
$$

$$
boxed{k = 284, mathrm{N/m}}
$$

Result

2 of 2

$$
k = 284, mathrm{N/m}
$$

Exercise 2

Step 1

1 of 2

Information given in the text are:

$k = 275, mathrm{N/m}$

$x = 14.3, mathrm{cm} = 0.143, mathrm{m}$

Force is given by:

$$
F = kx
$$

When we put known values into the previous equation we get:

$$
F = 275, mathrm{N/m} cdot 0.143, mathrm{m}
$$

$$
boxed{F = 39.3, mathrm{N}}
$$

Result

2 of 2

$$
F = 39.3, mathrm{N}
$$

Exercise 3

Solution 1

Solution 2

Step 1

1 of 2

Information given in the text are:

$m = 30.4, mathrm{g} = 0.0304, mathrm{kg}$

$x = 0.85, mathrm{m}$

We can determine spring constant by using Newton’s second law:

$$
kx = mg
$$

$g$ is gravitational acceleration which value is known $g = 9.81, mathrm{“m/s^{2}}$

$k$ will be:

$$
k = dfrac{mg}{x}
$$

When we put known values into the previous equation we get:

$$
k = dfrac{0.0304, mathrm{m} cdot 9.81, mathrm{m/s^{2}}}{0.85}
$$

$$
boxed{k = 0.35, mathrm{N/m} }
$$

Result

2 of 2

$$
0.35, mathrm{N/m}
$$

Step 1

1 of 2

We have:

$m g = k x$

Solve for k:

$$
k = dfrac{m g}{x} = dfrac{(0.0304)*(9.80)}{(0.85)} = 0.35 N/m
$$

Result

2 of 2

$$
0.35 N/m
$$

Exercise 4

Step 1

1 of 2

$E = (1/2) k (x_1 – x_2)^2$

$E = (1/2)*(350)*(0.85 – 0.050)^2$

$$
E = 112 N.m
$$

Result

2 of 2

$$
112 N.m
$$

Exercise 5

Step 1

1 of 2

$T = 2 pi sqrt{dfrac{l}{g}}$

$T^2 = (2 pi)^2 (dfrac{l}{g})$

$dfrac{l}{g} = dfrac{T^2}{(2 pi)^2}$

$$
l = dfrac{T^2 g}{(2 pi)^2}
$$

Result

2 of 2

$$
l = dfrac{T^2 g}{4 pi^2}
$$

Exercise 6

Solution 1

Solution 2

Step 1

1 of 2

Information given in the text are:

$T = 3, mathrm{s}$

Frequency is given by:

$$
f = dfrac{1}{T}
$$

When we put known values into the previous equation we get:

$$
f = dfrac{1}{3, mathrm{s}}
$$

$$
boxed{f = 0.3, mathrm{1/s} = 0.3, mathrm{Hz}}
$$

Result

2 of 2

$$
0.3, mathrm{Hz}
$$

Step 1

1 of 2

$$
f = dfrac{1}{T} = dfrac{1}{3} = 0.3 Hz
$$

Result

2 of 2

$$
0.3 Hz
$$

Exercise 7

Solution 1

Solution 2

Step 1

1 of 2

Waves: Identical

Direction: Same

Medium: Same

Result

2 of 2

A) Identical , Same , Same

Step 1

1 of 6

A standing wave can be produced by sending the wave to make one round-trip from your hand to the solid end. However, some requirements must be met for the standing wave to occur.

Step 2

2 of 6

**a)** The first option is that waves need to be identical, go in the same direction, and through the same medium. However, this is not correct since the waves won’t be in the same direction if the wave reflects off the solid end.

Step 3

3 of 6

**b)** In the second option, the waves need to be nonidentical, go in the opposite direction, and through a different medium. This answer is not correct.

Step 4

4 of 6

**c) If we want to produce standing waves, the waves need to be identical, going in the opposite direction and through the same medium. The wave then appears to be standing still, and because of that is called a standing wave.**

Step 5

5 of 6

**d)** The last option is for the waves to be nonidentical, go in the same direction, and through a different medium, but this won’t produce a standing wave.

Result

6 of 6

$$text{c)}$$

Exercise 8

Solution 1

Solution 2

Step 1

1 of 2

The speed of the wave:

$v = dfrac{2 d}{t} = dfrac{(2)*(11.2)}{4} = 5.6 m/s$

The frequency of the wave:

$$
f = dfrac{v}{lambda} = dfrac{5.6}{1.2} = 5 Hz
$$

Result

2 of 2

$$
5 Hz
$$

Step 1

1 of 5

**Given:**
– Wavelength: $d = 1.2 mathrm{~m}$;
– Distance: $d = 11.2 mathrm{~m}$;
– Time: $t = 4 mathrm{~s}$;

**Required:**

– The frequency $f$;

Step 2

2 of 5

Assume that the waves spread out at a constant speed given as the ratio of distance traveled and the time needed to travel that distance. We have an equation that relates the wave speed, wavelength and the frequency.
$$begin{align*}
v &= frac{s}{t}&&(1) \
v &= lambda f &&(2)
end{align*}$$

Step 3

3 of 5

First, we need to find the speed of the wave. Keep in mind that we are given the time needed for the wave to travel to the wall and back. That means that the distance we need to consider is twice the distance to the wall. Plugging the data into the first equation, we have:
$$begin{align*}
v &= frac{2d}{t} \
&= frac{2 cdot 11.2 mathrm{~m}}{4 mathrm{~s}} \
&= 5.6 ,frac{text{m}}{text{s}}
end{align*}$$

Step 4

4 of 5

Finally, the frequency can be found solving the second equation for $f$ dividing it by $lambda$:
$$begin{align*}
f &= frac{v}{lambda} \
&= frac{5.6 ,frac{text{m}}{text{s}} }{1.2 mathrm{~m}} \
&= 4.67 mathrm{~Hz} \
&approx 5 mathrm{~Hz}
end{align*}$$
$$boxed{ f approx 5 mathrm{~Hz} }$$

Result

5 of 5

$$f approx 5 mathrm{~Hz} $$

Exercise 9

Step 1

1 of 2

From the test $#5$ we have:

$l = dfrac{T^2 g}{(2 pi)^2}$

$l = dfrac{(4.89)^2 (9.80)}{((2) (3.1416))^2}$

$$
l = 5.94 m
$$

Result

2 of 2

$$
5.94 m
$$

Exercise 10

Step 1

1 of 2

We have:

$k x = m g$

Solve for k:

$k = dfrac{m g}{x}$

Where:

the units of m = [m] = kg

the units of g = [g] = $dfrac{m}{s^2}$

the units of x = [x] = meters = m

Thus:

$[k] = dfrac{[m] [g]}{[x]}$

$[k] = dfrac{(kg) (m/s^2)}{m}$

$[k] = dfrac{kg}{s^2}$

Result

2 of 2

$$
dfrac{kg}{s^2}
$$

unlock

All Solutions

Page 401: Section Review

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards