Question

A 121-cm-long, 4.00 g string oscillates in its m = 3 mode with a frequency of 180 Hz and a maximum amplitude of 5.00 mm. What are (a) the wavelength and (b) the tension in the string?

Answer 1

The wavelength of the string is 0.81 m and tension in the string is 14.69 N.

Step-by-step explanation:

To find the wavelength, we can use the formula: wavelength = 2L/n, where L is the length of the string and n is the mode number. In this case, the length of the string is 121 cm or 1.21 m, and the mode number is 3. So, the wavelength is 2(1.21)/3 = 0.81 m.

To find the tension in the string, we can use the formula: tension = μf^2λ, where μ is the linear mass density of the string, f is the frequency, and λ is the wavelength. Given μ = 4.00 g/m or 0.004 kg/m, f = 180 Hz, and λ = 0.81 m, we can plug in these values to find the tension: tension = (0.004)(180^2)(0.81) = 14.69 N.

Answer 2

0.8067 m

69.696 N

Step-by-step explanation:

m = Mode = 3

M = Mass of string = 4 g

f = Frequency = 180 Hz

l = Length of string = 121 cm

Length of the string is given by

`l=m(\lambda)/(2)\\\Rightarrow \lambda=2(l)/(m)\\\Rightarrow \lambda=2* (1.21)/(3)\\\Rightarrow \lambda=0.8067\ m`

The wavelength is 0.8067 m

Linear density is given by

`\mu=(M)/(l)\\\Rightarrow \mu=(4* 10^(-3))/(1.21)`

Speed of the wave

`v=f\lambda\\\Rightarrow v=180* 2* (1.21)/(3)\\\Rightarrow v=145.2\ m/s`

Speed of wave is given by

`v=\sqrt{(T)/(\mu)}\\\Rightarrow T=\mu v^2\\\Rightarrow T=(4* 10^(-3))/(1.21)* 145.2^2\\\Rightarrow T=69.696\ N`

Tension is given by 69.696 N