Oscillations and Waves Questions and Answers

Answer: C

The speed of sound in air does not depend on the Pressure of air.

For an ideal gas the speed of sound is

\[ \begin{aligned} v &= \sqrt{\frac{\gamma P}{\rho}} \\ &= \sqrt{\frac{\gamma R T}{M}} \end{aligned} \]

If the pressure rises at constant temperature, the density rises in the same proportion, so the ratio \(P/\rho\) and the speed stay the same. The speed therefore depends on temperature, not on pressure.

Density matters when it changes independently, and temperature raises the speed by about 0.6 m/s per degree Celsius. Humidity matters too: moist air is less dense than dry air at the same temperature and pressure, so sound travels slightly faster in it.

Answer: A

On the moon the period is longer: \(T_{m} \gt T_{e}\).

A simple pendulum's period is T = \(2\pi\sqrt{L/g}\). The moon's surface gravity is about one-sixth of the earth's, so:

\[ \begin{aligned} T_{m} &= T_{e}\sqrt{\frac{g}{g/6}} \\ &= \sqrt{6}\,T_{e} \approx 2.45\,T_{e} \end{aligned} \]

The period would be equal only if g were the same, and shorter only if gravity were stronger, so the other options are wrong.

Answer: A

The period is \(2\pi\sqrt{\frac{m}{k_{1}+k_{2}}}\).

Springs in parallel stretch by the same amount, so their forces add and the effective spring constant is k₁ + k₂. The period of a mass-spring system is \(2\pi\sqrt{m/k}\), which gives option A.

Option B is the period for springs in series, where k_eff = \(\frac{k_{1}k_{2}}{k_{1}+k_{2}}\). Option C subtracts the constants, which no arrangement does, and option D inverts the series result.