The speed of a transverse wave on a stretched string is given by the formula v = sqrt((T)/(mu)), where T is the tension in the string and mu is the linear mass density (mass per unit leng

The speed of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the string and $\mu$ is the linear mass density (mass per unit length) of the string. Both tension and linear density are properties of the medium that influence the speed of the wave. Wavelength and frequency are related to the speed ($v = f\lambda$) but do not determine the speed itself; rather, they are determined by the speed and each other.

Therefore, both Tension (A) and Linear density of the string (D) are factors that influence the speed of the wave. Since this is a multiple-choice question and typically only one answer is expected, we will select the first correct option listed.

The correct option is A.
$\boxed{\text{A. Tension}}$

The sound intensity level $\beta$ in decibels is related to the sound intensity $I$ by the formula:

$$\beta = 10 \log_{10}\left(\frac{I}{I_0}\right)$$

where $I_0$ is the reference intensity, which is the threshold of human hearing, $I_0 = 1.0 \times 10^{-12}\,W/m^2$.

The speed of sound in an ideal gas is given by the formula:

$$v = \sqrt{\frac{\gamma RT}{M}}$$

where $\gamma$ is the adiabatic index, $R$ is the ideal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.
From this formula, we can see that the speed of sound $v$ is directly proportional to the square root of the absolute temperature $T$:

$$v \propto \sqrt{T}$$

This problem involves the Doppler effect for sound. When an observer approaches a stationary source, the observed frequency $f'$ is given by the formula:

$$f' = f \left(\frac{v + v_o}{v}\right)$$

where $f$ is the source frequency, $v$ is the speed of sound in the medium, and $v_o$ is the speed of the observer.

For a standing sound wave in a tube:

• At a closed end, the air molecules cannot move, so there is no displacement. This point is a displacement node.
• At an open end, the air molecules are free to move with maximum amplitude. This point is a displacement antinode.

Therefore, a standing sound wave in a tube opened at one end has a displacement node at the closed end and a displacement antinode at the open end.

The correct option is B.
$\boxed{\text{B. node, antinode}}$