Question

practice

1. if a 2.5 m long string on the same wave machine has a tension of 240 n, and the wave speed is 300 m/s, what is the mass of the string? t/i ans: 6.7 × 10⁻³ kg

1. if a wave machine string has a linear density of 0.2 kg/m and a wave speed of 200 m/s, what tension is required? t/i ans: 8 × 10³ n

1. if a string on a wave machine has a linear density of 0.011 kg/m and a tension of 250 n, what is the wave speed? t/i ans: 1.5 × 10² m/s

Explanation:

Problem 1

Step1: Recall wave speed formula

The formula for the speed \( v \) of a wave on a string is \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is tension and \( \mu \) is linear mass density (\( \mu=\frac{m}{L} \), \( m \) is mass, \( L \) is length). Rearranging for \( m \): first find \( \mu \) from \( v = \sqrt{\frac{T}{\mu}} \Rightarrow \mu=\frac{T}{v^2} \), then \( m=\mu L \).

Step2: Calculate linear density \( \mu \)

Given \( T = 240\space N \), \( v = 300\space m/s \), so \( \mu=\frac{240}{(300)^2}=\frac{240}{90000}=\frac{2}{750}\space kg/m \).

Step3: Calculate mass \( m \)

Length \( L = 2.5\space m \), so \( m=\mu L=\frac{2}{750}\times2.5=\frac{5}{750}=\frac{1}{150}\approx 6.7\times 10^{-3}\space kg \).

Step1: Recall wave speed formula

Use \( v = \sqrt{\frac{T}{\mu}} \), rearrange for tension \( T \): \( T = \mu v^2 \).

Step2: Substitute values

Given \( \mu = 0.2\space kg/m \), \( v = 200\space m/s \), so \( T = 0.2\times(200)^2=0.2\times40000 = 8000 = 8\times 10^{3}\space N \).

Step1: Recall wave speed formula

Use \( v = \sqrt{\frac{T}{\mu}} \).

Step2: Substitute values

Given \( T = 250\space N \), \( \mu = 0.011\space kg/m \), so \( v=\sqrt{\frac{250}{0.011}}\approx\sqrt{22727.27}\approx 150.76\approx 1.5\times 10^{2}\space m/s \).

Answer:

\( 6.7\times 10^{-3}\space kg \)

Problem 2