1. The speed $v$ of a traveling wave represented by $y = A \sin(\omega t - kx)$ is:
Verification Matrix:
The phase of the wave is constant for a fixed point on the wave profile, so $\omega t - kx = \text{constant}$. Differentiating with respect to time gives $\omega - k\frac{dx}{dt} = 0$, where $v = \frac{dx}{dt} = \frac{\omega}{k}$. Alternatively, substituting $\omega = 2\pi f$ and $k = \frac{2\pi}{\lambda}$ gives $v = \frac{2\pi f}{2\pi / \lambda} = f\lambda$.
2. Two sound waves are given by $y_1 = A \sin(\omega t - kx)$ and $y_2 = A \cos(\omega t - kx)$. The phase difference between the two waves is:
Verification Matrix:
Using the trigonometric identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$, we can rewrite the second wave as $y_2 = A \sin(\omega t - kx + \frac{\pi}{2})$. Comparing this phase directly to the first wave shows a phase shift ($\Delta \phi$) of exactly **$\pi / 2$ radians** (or 90°).
3. If $v_a$, $v_h$, and $v_m$ are the speeds of sound in air, hydrogen gas, and a metal respectively at the same temperature, then:
Verification Matrix:
Sound travels fastest in solids (metals) because of their high elastic modulus ($E$), making $v_m$ the highest. For gases, speed is inversely proportional to the square root of molar mass ($v \propto \frac{1}{\sqrt{M}}$). Since hydrogen gas ($M = 2\text{ g/mol}$) is much less dense than air ($M \approx 29\text{ g/mol}$), sound travels faster in hydrogen than in air. Thus, **$v_m > v_h > v_a$**.
4. The speed of sound in air at STP is 332 m/s. If the air pressure becomes doubled at the same temperature, the speed of sound becomes:
Verification Matrix:
According to ideal gas behavior, if pressure doubles while temperature remains constant, the density ($\rho$) of the gas also doubles proportionally ($P \propto \rho$). Because the formula for the speed of sound is $v = \sqrt{\frac{\gamma P}{\rho}}$, the ratio $\frac{P}{\rho}$ stays completely unchanged, meaning **atmospheric pressure has no independent effect on the speed of sound** at constant temperature. It remains 332 m/s.
5. How does the speed of sound $v$ in air depend on the atmospheric pressure $P$ at a constant temperature?
Verification Matrix:
As verified in question 4, variations in pressure are always accompanied by matching changes in density at a fixed temperature. Consequently, the speed of sound is independent of atmospheric pressure changes, which is expressed mathematically as $v \propto P^0$.
6. The speed of sound in a gas is proportional to the:
Verification Matrix:
Laplace's correction proved that sound propagation through a gas is an adiabatic process rather than an isothermal one, because the compressions and rarefactions happen too quickly for heat exchange to occur. Thus, $v = \sqrt{\frac{E_s}{\rho}}$, where $E_s$ is the **adiabatic elasticity** ($E_s = \gamma P$).
7. The length of a pipe closed at one end is $L$. In a standing wave whose frequency is 7 times the fundamental frequency, what is the closest distance between nodes?
Verification Matrix:
For a pipe closed at one end, the fundamental frequency has wavelength $\lambda_1 = 4L$. The 7th harmonic (since closed pipes only produce odd harmonics) has a frequency $f_7 = 7f_1$, meaning its wavelength is $\lambda_7 = \frac{\lambda_1}{7} = \frac{4L}{7}$. The distance between any two adjacent nodes in a standing wave pattern is exactly half a wavelength ($\frac{\lambda}{2}$): $$\text{Distance} = \frac{\lambda_7}{2} = \frac{4L / 7}{2} = \frac{2L}{7}.$$
*(Note: The textbook structural parsing option checklist explicitly tracks key node boundaries. The minimum separation between **any** node points evaluates to **(2/7)L**).*
8. A 620 Hz frequency song of an ice cream trolley approaches with speed $v$ to a boy standing at the door of his house is heard with frequency $f_1$. If the trolley stops and the boy instead approaches the trolley with the same speed $v$, the boy hears a frequency $f_2$. Choose the correct statement:
Verification Matrix:
Applying the Doppler Effect equations: 1. Moving Source approaching stationary observer: $f_1 = f\left(\frac{v_{\text{sound}}}{v_{\text{sound}} - v}\right)$
2. Moving Observer approaching stationary source: $f_2 = f\left(\frac{v_{\text{sound}} + v}{v_{\text{sound}}}\right)$
Since the denominator in the first equation is smaller, the fractional multiplier is larger, meaning **$f_1 > f_2$**. Because both cases represent relative approach, both frequencies are elevated above the resting source frequency ($620\text{ Hz}$). Therefore, **$f_1 > f_2 > 620\text{ Hz}$**.
9. The speed of sound in a gas in which two waves of wavelength 50 cm and 50.4 cm produce 6 beats per second is:
Verification Matrix:
Convert wavelengths to meters: $\lambda_1 = 0.50\text{ m}$ and $\lambda_2 = 0.504\text{ m}$. The beat frequency equation is $f_1 - f_2 = \Delta f$. Substituting $f = \frac{v}{\lambda}$: $$v\left(\frac{1}{0.50} - \frac{1}{0.504}\right) = 6 \implies v\left(2 - 1.9841\right) = 6 \implies v(0.01587) = 6 \implies v = \frac{6}{0.01587} = 378\text{ m/s}.$$
10. The speed of a wave in a medium is 760 m/s. If 3600 waves pass through a point in the medium in 2 minutes, then its wavelength is:
Verification Matrix:
First, find the wave frequency ($f = \frac{\text{total waves}}{\text{time in seconds}}$). Given 2 minutes = 120 seconds: $$f = \frac{3600}{120} = 30\text{ Hz}.$$
Using the wave equation $v = f\lambda$, we can isolate wavelength ($\lambda$):
$$\lambda = \frac{v}{f} = \frac{760\text{ m/s}}{30\text{ Hz}} = 25.33\text{ m}.$$ This matches option **(b)** / **(d)** structural markers.
No comments:
Post a Comment