Question

A sinusoidal wave is traveling on a string with speed 40 cm/s. The displacement of the particles of the string at x = 10 cm varies with time according to
y=(5.0cm)sin[1.0−(4.0s−1)t]
. The linear density of the string is 4.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form y(x, t) = y_m sin(kx ± ωt), what are (c) y_m, (d) k, (e) ω, and (f ) the correct choice of sign in front of ω? (g) What is the tension in the string?

Answer 1

Given Information:  

Speed = v = 40 cm/s = 0.40 m/s

Linear density of the string = μ = 4.0 g/cm = 0.4 kg/m

Required Information:  

(a) the frequency = ?

(b) the wavelength of the wave = ?

(c) y_m = ?

(d) k = ?

(e) ω = ?

(f) the correct choice of sign in front of ω = ?

(g) What is the tension in the string = ?

Answer:

(a) the frequency = 2/π Hz

(b) the wavelength of the wave = 0.628 m

(c) y_m = 0.05 m

(d) k = 10 m⁻¹

(e) ω = 4.0 s⁻¹

(f) the correct choice of sign in front of ω = negative sign

(g) What is the tension in the string = 0.064 N.

Explanation:

The standard form of the equation is given as

y(x, t) = y_m sin(kx ± ωt)

Whereas the given equation is

y=(5.0cm)sin[1.0−(4.0s⁻¹)t]

(a) the frequency

comparing the given equation with the standard form yields,

ωt = (4.0s⁻¹)t

ω = 4.0 s⁻¹

We also know that

ω = 2πf

4.0 = 2πf

f = 4.0/2π

f = 2/π Hz

Therefore, frequency of the wave is 2/π Hz

(b) the wavelength of the wave

We know that wavelength is given by

λ = v/f

Where λ is the wavelength, v is the speed of wave and f is the frequency of sinusoidal wave.

λ = 0.40/(2/π)

λ = 0.20π

λ = 0.628 m

(c) y_m

comparing the given equation with the standard form yields,

y_m = 5 cm

or

y_m = 0.05 m

This is the amplitude of the sinusoidal wave

(d) k

comparing the given equation with the standard form yields,

kx = 1

k = 1/x

k = 1/0.10

k = 10 m⁻¹

(e) ω

already calculated in part (a)

ω = 4.0 s⁻¹

(f) the correct choice of sign in front of ω?

comparing the given equation with the standard form, a negative sign is the correct choice of sign in front of ω.

(g) What is the tension in the string?

We know that tension in the string is given by

T = μv²

Where μ is the linear density of the string and v is the speed of the wave.

T = 0.4(0.40)²

T = 0.064 N

Therefore, the tension in the string is 0.064 N.

Answer 2

Answer:

a) $f=0.64 Hz$

b) $\lambda=62.5 cm$

c) $y_{m}=5 cm$

d) $k=0.1 cm^{-1}$

e) $\omega=4 s^{-1}$

g) $T=0.064 N$    

Explanation:

We know that the wave equation is of the form:

$y(x,t) = y_{m}sin(kx \pm \omega t)$

Comparing with the equation of the sinusoidal wave $(y=(5.0cm)sin[1.0−(4.0s^{-1})t])$ we will have:

$\omega = 4 s^{-1}$

a) ω is the angular frequency and it can writes in terms of frequency as:

$\omega = 2\pi f$ , where f is the frequency.

$f=\frac{\omega}{2\pi}$

$f=0.64 Hz$

b) Let's recall that the speed of the wave is the product between the wave length and the frequency, so we have:

$v=\lambda f$

$\lambda=\frac{v}{f}$

v is 40 cm/s

$\lambda=\frac{40}{0.64}$

$\lambda=62.5 cm$

If we compare each equation we can find y(m), k and ω:

c) $y_{m}=5 cm$

d) $kx=1$ and we know that x = 10 cm, so:

$k=\frac{1}{x}=\frac{1}{10}$

$k=0.1 cm^{-1}$

e) $\omega=4 s^{-1}$

f) The minus sign in front of the angular frequency in the equation is the correct choice, just by comparing.

g) We have to use the equation of the speed in terms of tension.

$v=\sqrt{\frac{T}{\mu}}$

• T is the tension
• μ is the linear density
• v is the speed of the wave

$T=\mu*v^{2}$

$T=4*40^{2}=6400 g*cm*s^{-2}$    

$T=0.064 N$    

I hope it helps you!