Question

Write the equation of a wave traveling along the +x-axis with an amplitude of 0.02 m, a frequency of 440 Hz, and a speed of 330m/s.

Answer 1

The equation of the wave traveling along the +x-axis with the given amplitude, frequency, and speed is y(x, t) = 0.02 sin(((880π)/330) x - (880πt)).

Step-by-step explanation:

• The equation of a wave can be expressed as y(x, t) = A sin(kx - ωt), where 'y' = displacement, 'A' = amplitude, 'k' = wavenumber, 'x' = position, 'ω' = angular frequency, and 't' = time.
• To find the equation for the wave described, we first calculate the wavenumber 'k' using the formula k = ω/v where 'v' is the wave speed. The angular frequency 'ω' is given by ω = 2πf, with 'f' being the frequency.

For a wave traveling along the +x-axis with an amplitude of 0.02 m, a frequency of 440 Hz, and a speed of 330 m/s, the angular frequency 'ω' is (2π × 440) rad/s, and the wavenumber 'k' is (2π × 440)/330 m-1. Therefore, the required wave equation is:

• y (x, t) = 0.02 sin(((2π × 440)/330)x - (2π × 440)t)

• y (x, t) = 0.02 sin(((880π)/330) x - (880πt)).

Answer 2

The equation of the wave travelling along the +x-axis is y = 0.02 sin (880π/330 x – 880 πt)

Step-by-step explanation:

Given data

Amplitude 0.02 m , Frequency= 440 Hz ,Speed = 330 m/s

The equation format is written as,

y = A sin ( k x – ω t)

We need the value of A, k, x -ω t

1. Find the k value

v = f ×λ

330 = 440×λ

k = 2π×λ

k = 880 π /330 m-1

440 ×2π = w

2. Find the ω value

f×2π =ω

ω = 880 π s-1

3. Find the A value

we get A value from the given data

A = 0.02 m

By the formula,

y = A sin ( k x – ω t)

Substitute the values we get the equation,

y = 0.02 sin (880π/330 x – 880 πt)

The equation of the wave travelling along the +x-axis is y = 0.02 sin (880π/330 x – 880 πt)