Question

On a distant planet where the gravitational acceleration is 5.52 m/s^2, an astronaut hangs a 112-gram ball from the end of a wire. She observes the speed of transverse pulse along the wire to be 44.4 m/s. The linear density of the wire is closest to: (a) 0.131 g/m (b) 0.172 g/m (c) 0.227 g/m (d) 0.314 g/m (e) 0.462 g/s

Answer 1

The linear density of the wire is 0.314 g/m.

Step-by-step explanation:

It is given that,

Acceleration,
`a=5.52\ m/s^2`

Mass of the ball, m = 112 gm

Speed of the transverse wave, v = 44.4 m/s

The speed of the transverse wave is given by :

`v=\sqrt{(T)/(\mu)}`

Where

T = tension in the wire

`\mu` = mass per unit length

`\mu=(T)/(v^2)`

`\mu=(ma)/(v^2)`

`\mu=(112\ g* 5.52\ m/s^2)/((44.4\ m/s)^2)`

`\mu=0.3136\ g/m`

or

`\mu=0.314\ g/m`

So, the linear density of the wire is 0.314 g/m. Hence, this is the required solution.