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Corrected Question:
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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
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Answer:
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(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
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Explanation:
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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)
<em>Where;</em>
k = 628.3/cm
<em>Substitute k = 628.3/cm into equation (iii) as follows;</em>
628.3 = 2π / λ [Take π = 3.142]
<em>Solve for λ;</em>
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)
<em>where;</em>
w = 6283/s
<em>Substitute w = 6283/s into equation (iv) as follows;</em>
6283 = 2 π f [Take π = 3.142]
<em>Solve for f;</em>
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).
