The arrangement in the drawing shows a block (mass = 15.0 kg) that is held in po

Question

The arrangement in the drawing shows a block (mass = 15.0 kg) that is held in position on a frictionless incline by a cord (length = 0.595 m). The mass per unit length of the cord is 1.22e-2 kg/m, so the mass of the cord is negligible compared to the mass of the block. The cord is being vibrated at a frequency of 162 Hz (vibration source not shown in the drawing). What is the smallest angle ? between 15.0degree and 90.0degree at which a standing wave exists on the cord?

location: https://general.physics.rutgers.edu/gifs/CJ/17-15.gif

Explanation / Answer

The mass of the block M = 15kg

the linear mass density = 1.22*10^-2 kg/m

the length of the string L = 0.595m

The frequency of vibration f = 162Hz

Then from Newton;s law

   tension   T = Mgsin

Now the frequency

             f = 1/2L T/

then     f^2 = 1/4L^2 (T/)

            Mgsin = 4 f^2 L^2

                 sin = 4(162)^2(0.595)^2(1.22*10^-2) / (15)(9.8)

solve for angle

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