A string with a mass per length of 2.00 g/m is stretched between two points that are 0.400 m apart.The third mode frequency of the stretched string is 20.0 Hz lower than the fourth mode frequency.What is the tension in the string?
A) 180 N
B) 160 N
C) 140 N
D) 120 N
E) 100 N

Question

Quizplus · Accepted Answer

The tension in the string can be calculated using the formula:$$f_n = \frac{nv}{2L}$$where:$$f_n$$ is the frequency of the nth mode$$n$$ is the mode number$$v$$ is the speed of the wave on the string$$L$$ is the length of the stringThe speed of the wave on the string is given by:$$v = \sqrt{\frac{T}{\mu}}$$where:$$T$$ is the tension in the string$$\mu$$ is the mass per length of the stringSubstituting the given values into the above equations, we get:$$f_3 = \frac{3v}{2L}$$$$f_4 = \frac{4v}{2L}$$$$v = \sqrt{\frac{T}{\mu}}$$Dividing the first equation by the second equation, we get:$$\frac{f_3}{f_4} = \frac{3}{4}$$Substituting the given values into the above equation, we get:$$\frac{20.0	ext{ Hz}}{f_4} = \frac{3}{4}$$Solving for $$f_4$$, we get:$$f_4 = 26.67	ext{ Hz}$$Substituting the given values and the calculated value of $$f_4$$ into the first equation, we get:$$20.0	ext{ Hz} = \frac{3v}{2(0.400	ext{ m})}$$Solving for $$v$$, we get:$$v = 13.33	ext{ m/s}$$Substituting the given values and the calculated value of $$v$$ into the equation for the speed of the wave on the string, we get:$$13.33	ext{ m/s} = \sqrt{\frac{T}{2.00	imes10^{-3}	ext{ kg/m}}}$$Solving for $$T$$, we get:$$T = 180	ext{ N}$$Therefore, the tension in the string is 180 N.

A String with a Mass Per Length of 2