The variation in the pressure of helium gas,measured from its equilibrium value,is given by $\Delta$P = 2.9 *10 ^- ⁵ cos (6.20x - 3 000t)where x and t have units m and s,and $\Delta$P is measured in N/m².Determine the frequency (in Hz)of the wave. A)1 500 B)477 C)1.01 D)0.32 E)239

Question

Quizplus · Accepted Answer

The equation for the pressure variation is of the form:
P = A cos (kx - ωt)
where A is the amplitude, k is the wave number, ω is the angular frequency, and x and t have their usual meanings. The frequency of the wave is given by:
f = ω/2π
Comparing this with the given equation, we have:
k = 6.20 x 10^3 m^-1
ω = 2πf = 2π(3.00 x 10^3) rad/s
So, the frequency of the wave is:
f = ω/2π = (3.00 x 10^3)/(2π) Hz ≈ 477 Hz. 
Therefore, the best choice is (B) 477 with the explanation above.

The Variation in the Pressure of Helium Gas,measured from Its Δ\DeltaΔ

The Variation in the Pressure of Helium Gas,measured from Its $\Delta$