Question

A transverse traveling wave on a taut wire has an amplitude of 0.200 mm and a frequency of 500 Hz. It travels with a speed of 196 m/s. (a) Write an equation in SI units of the form y 5 A sin (kx 2 vt) for this wave. (b) The mass per unit length of this wire is 4.10 g/m. Find the tension in the wire.

Answer 1

(a) y = (0.2 x 10^-3 m)Sin (16 x - 3140 t)

(b) 157.5 N

Step-by-step explanation:

Amplitude, A = 0.2 mm

Frequency, f = 500 Hz

velocity,v = 196 m/s

(a) The standard equation of a wave is

y = A Sin (kx - ωt)

Where, k = 2π/λ

where, λ is the wavelength

λ = v / f = 196 / 500 = 0.392 m

So, k = 2 x 3.14 / 0.392 = 16

ω = 2 x π x f = 2 x 3.14 x 500 = 3140 rad/s

So, the equation is

y = (0.2 x 10^-3 m)Sin (16 x - 3140 t)

(b) mass per unit length, m = 4.10 g/m = 4.10 x 10^-3 kg/m

The velocity

`v=\sqrt{(T)/(m)}`

T = v² x m

T = 196 x 196 x 4.10 x 10^-3 = 157.5 N

Thus, the tension in the wire is 157.5 N.

Answer 2

(a). The value of A, k and ω is
`2*10^(-4)\ m`, 16.02 rad/m and 3141.59 rad/s.

(b). The tension in the wire is 157.5 N.

Step-by-step explanation:

Given that,

Amplitude = 0.200 mm

Frequency = 500 Hz

Speed = 196 m/s

Mass per unit length = 4.10 g/m

Suppose we need to calculate the parameters A, k, and ω.

The equation given by

`y=A\sin(kx-\omega t)`

(a). We need to calculate the amplitude

Using formula of amplitude

`A=0.200*10^(-3)\ m`

`A=2*10^(-4)\ m`

We need to calculate the angular frequency

Using formula of angular frequency

`\omega=2\pi f`

`\omega=2*\pi*500`

`\omega=3141.59\ rad/sec`

We need to calculate the angular wave number

Using formula of angular wave number

`k=(\omega)/(v)`

Put the value into the formula

`k=(3141.59)/(196)`

`k=16.02\ rad/min`

(b). We need to calculate the tension in the wire

Using formula of tension in the wire

`T=v^2*\mu`

Put the value into the formula

`T=196^2*4.10*10^(-3)`

`T=157.5\ N`

Hence, (a). The value of A, k and ω is
`2*10^(-4)\ m`, 16.02 rad/m and 3141.59 rad/s.

(b). The tension in the wire is 157.5 N.