The speed of the wave when it is stretched by 0.9 m is 9 m/s.
The given parameters:
- Original length of the spring, l = 0.3 m
- Length of the spring when stretched = 0.6 m
- Speed of the wave, v = 4.5 m/s
The extension of the spring when it is stretched by 0.6 m calculated as follows;
x₁ = 0.6 m - 0.3 m
x₁ = 0.3 m
The extension of the spring when it is stretched by 0.9 m calculated as follows;
x₂ = 0.9 m - 0.3 m
x₂ = 0.6 m
The speed of the wave for the second extension can be determined by applying the principle of conservation of energy;
[tex]\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\\frac{v^2}{x^2} = \frac{k}{m} \\\\\frac{v}{x} = \sqrt{\frac{k}{m} } \\\\\frac{v_1}{x_1} = \frac{v_2}{x_2}\\\\v_2 = \frac{v_1x_2}{x_1} \\\\v_2 = \frac{4.5 \times 0.6}{0.3} \\\\v_2 = 9 \ m/s[/tex]
Learn more about conservation of energy here: https://brainly.com/question/166559
