For a harmonic wave given by determine (a) wavelength; (b) frequency; (c) propagation constant; (d) angular frequency; (e) period; (f) velocity; (g) amplitude.

For a harmonic wave given by y = (10 cm) sin[(628.3/cm)x − (6283/s)t]

determine (a) wavelength; (b) frequency; (c) propagation constant; (d) angular frequency; (e) period; (f) velocity; (g) amplitude.

Source: https://www.chegg.com/homework-help/harmonic-wave-given-y-10-cm-sin-6283-cm-x-6283-s-t-determine-chapter-4-problem-7p-solution-9780131499331-exc

(a)  wavelength = 0.01m

(b) frequency = 999.84Hz

(c) propagation constant =  628.3/cm

(d) angular frequency = 6283/s

(e) period = 0.001s

(f) velocity = 9.9984m/s

(g) amplitude = 10cm

The general equation of a simple harmonic wave motion is given by;

y(x, t) = A sin (kx ± wt)           (i)

where;

y = vertical displacement

A = Amplitude

k = wave number = 2 π /  λ   [λ = wavelength]

w = angular velocity = 2π / T or  2πf  [T = period of oscillation, f = frequency]

t = time taken for displacement

x = horizontal displacement

The given equation is;

y = (10 cm) sin[(628.3/cm)x − (6283/s)t]        (ii)

Comparing the equations (i) and (ii)

y(x, t) = A sin (kx ± wt)           (i)

y = (10 cm) sin[(628.3/cm)x − (6283/s)t]        (ii)  

From comparison;

A = 10cm

k = 628.3/cm

w = 6283/s

(a) To calculate the wavelength, λ, we use the wave number relation as follows;

k = 2 π /  λ    (iii)

Where;

k = 628.3/cm

Substitute k = 628.3/cm into equation (iii) as follows;

628.3 = 2π / λ       [Take π = 3.142]

Solve for λ;

λ = 2(3.142) / 628.3

λ = 6.284 / 628.3

λ = 0.01m

Therefore the wavelength is 0.01m

(b) To calculate the frequency, f, we use;

w = 2π f          (iv)

where;

w = 6283/s

Substitute w = 6283/s into equation (iv) as follows;

6283 = 2 π f           [Take π = 3.142]

Solve for f;

f = 6283 / (2 π)

f = 6283 / (2 x 3.142)

f = 6283 / (6.284)

f = 999.84 Hz

Therefore, the frequency is 999.84 Hz

(c) The propagation constant is also called the wave number k.

As shown in the comparison of equations i and ii,

k = 628.3/cm

(d) The angular velocity, w = 6283/s     [as shown in the comparison of equations i and ii]

(e) To calculate the period, T, we use;

T = 1 / f       (v)

i.e the inverse of the frequency gives the period

Where;

f = 999.84 Hz

Substitute f = 999.84Hz into equation (v)

T = 1 / 999.84

T = 0.001 s

Therefore the period is 0.001 s

(f) The velocity, v, is the product of the wavelength (λ) and the frequency (f) as follows;

v = f x λ

Where;

f = 999.84 Hz

λ = 0.01m

Substitute these values into equation (vi) as follows;

v = 999.84 x 0.01

v = 9.9984m/s

Therefore, the velocity is 9.9984m/s

(g) The amplitude, A, is 10cm as shown in the comparison of equations (i) and (ii).

i don’t know what the answer is i wish i

area of focus