An object of mass m = 5.0 kg hangs from a cord around a light pulley: the length of the cord between the oscillator and the pulley is l = 2.0 m. when the oscillator is set to a frequency of 150 hz, a standing wave with six loops is formed. what must be the linear mass density of the cord?

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

Where is the length of the string and the velocity of propagation. Use this expression to find the value of .

The velocity of propagation is given by the expression:

Where is the desirable variable of the problem, the linear mass density, and is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

With the value of the tension and the velocity you can find the mass density:

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force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2