Chapter 16

(a) $1\text{.}\text{23}\times {\text{10}}^3\text{N/m}$

(b) $6\text{.}\text{88 kg}$

(c) $4\text{.00 mm}$

(a) 889 N/m

(b) 133 N

(a) $6\text{.}\text{53}\times {\text{10}}^3\text{N/m}$

(b) Yes

16.7 ms

$\mathrm{0.400\; s}/\text{beats}$

400 Hz

12,500 Hz

1.50 kHz

(a) 93.8 m/s

(b) $\text{11}\text{.}3\times {\text{10}}^3\text{rev/min}$

$2\text{.}\text{37}\text{N/m}$

0.389 kg

94.7 kg

1.94 s

6.21 cm

2.01 s

2.23 Hz

(a) 2.99541 s

(b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by $(0\text{.}\text{01}{)}^2=0\text{.}\text{01\%}$ so it is necessary to have at least 4 digits after the decimal to see the changes.

(a) Period increases by a factor of 1.41 ($\sqrt{2}$)

(b) Period decreases to 97.5% of old period

Slow by a factor of 2.45

length must increase by 0.0116%.

(a) $\text{1.99 Hz}$

(b) 50.2 cm

(a) $3\text{.}\text{95}\times {\text{10}}^6\text{N/m}$

(b) $7\text{.}\text{90}\times {\text{10}}^6\text{J}$

a). 0.266 m/s

b). 3.00 J

$\pm \frac{\sqrt{3}}{2}$

384 J

(a). 0.123 m

(b). −0.600 J

(c). 0.300 J. The rest of the energy may go into heat caused by friction and other damping forces.

(a) $5\text{.}\text{00}\times {\text{10}}^5\text{J}$

(b) $\text{1.20}\text{×}{\text{10}}^3$ s

$t=9\text{.}\text{26 d}$

$f=\text{40}\text{.}\text{0 Hz}$

$v_{\text{w}}=\text{16}\text{.}\text{0 m/s}$

$\lambda =\text{700 m}$

$d=\text{34}\text{.}\text{0 cm}$

$f=\text{4 Hz}$

462 Hz,

4 Hz

(a) 3.33 m/s

(b) 1.25 Hz

0.225 W

7.07

16.0 d

2.50 kW

$\text{3.38}\times {\text{10}}^{\text{–5}}{\text{W/m}}^2$

(d)

(b)

The frequency is given by

$$f=\frac{1}{T}=\frac{50cycles}{30s}=1.66Hz$$

Time period is:

$$T=\frac{1}{f}=\frac{1}{1.66}=0.6\text{s}$$

(c)

The energy of the particle at the center of the oscillation is given by

$$E=\frac{1}{2}m{v}^2$$ $$=\frac{1}{2}\times 0.2\text{kg}\times {(5{\text{ m·s}}^{-1})}^2$$ $$=2.5\text{J}$$

(b)

19.7 J

(c)

$$d=\frac{k}{2{\mu }_{K}mg}{\left(X^2-\frac{{\mu }_{K}mg}{k}\right)}^2$$ $$wherek=50\text{N}\cdot {\text{m}}^{-1}$$ $${\mu }_k=0.06$$ $$m=0.5\text{kg}$$ $$d=\frac{50\text{N}\cdot {\text{m}}^{-1}}{2\times 0.06\times 9.8\text{m}\cdot {\text{s}}^{-2}}({(0.2)}^2-(\frac{{(0.06\times 0.5\text{kg}\times 9.8\text{m}\cdot {\text{s}}^{-2})}^2}{{(50\text{N}\cdot {\text{m}}^{-1})}^2}\left)\right)$$ $$=1.698\text{m}$$

The waves coming from a tuning fork are mechanical waves that are longitudinal in nature, whereas electromagnetic waves are transverse in nature.

The sound energy coming out of an instrument depends on its size. The sound waves produced are relative to the size of the musical instrument. A smaller instrument such as a tambourine will produce a high-pitched sound (higher frequency, shorter wavelength), whereas a larger instrument such as a drum will produce a deeper sound (lower frequency, longer wavelength).

$2\pi \text{m}$

The student explains the principle of superposition and then shows two waves adding up to form a bigger wave when a crest adds with a crest and a trough with another trough. Also the student shows a wave getting cancelled out when a crest meets a trough and vice versa.

The student must note that the shape of the wave remains the same and there is first an overlap and then receding of the waves.

(c)