Question

one end of a horizontal string is attached to a small amplitude mechanical 60 Hz oscillator. The string's mass per unit length is 3.5 x 10⁻⁴ kg/m. The string passes over a pulley, a distance l= 1.5 m away, and weights are hung from this end. What mass m must be hung from this end of the string to produce one loop.

Answer 1

The mass m that must be hung from the end of the string to produce one loop is approximately 1.04 kg.

Step-by-step explanation:

The mass m that must be hung from the end of the string to produce one loop depends on the frequency of the oscillator and the linear density of the string. In this case, the oscillator has a frequency of 60 Hz, and the linear density of the string is given as 3.5 x 10⁻⁴ kg/m. To find the mass, we can use the formula:

m = (f² * l) / (4π² * μ)

where f is the frequency of the oscillator, l is the distance to the pulley, and μ is the linear density of the string. Plugging in the values, we get:

m = (60² * 1.5) / (4π² * 3.5 x 10⁻⁴)

Simplifying the calculation, the mass m that must be hung from the end of the string to produce one loop is approximately 1.04 kg.