Question

On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced had a frequency of 120 Hz and wavelength of 6 cm. (a) What was the tension in the cord? (b) How large a mass M must be hung from its end to give it this tension? ​

Answer 1

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Step-by-step explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

`v =f\lambda`

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

`\mu = (m)/(l)`

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

`\mu = (1.45\ x\ 10^(-3)\ kg)/(5\ m)\\\\\mu = 2.9\ x\ 10^(-4)\ kg/m`

Now, for the tension we use the formula:

`v = \sqrt{(T)/(\mu)}\\\\7.2\ m/s = \sqrt{(T)/(2.9\ x\ 10^(-4)\ kg/m)}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^(-4)\ kg/m) = T`

T = 0.015 N

(b)

The mass to be hung is:

`T = Mg\\\\M = (T)/(g)\\\\M = (0.015\ N)/(9.8\ m/s^2)\\\\`

M = 1.53 x 10⁻³ kg = 1.53 g