Question

g 2. In a laboratory experiment on standing waves a string 3.0 ft long is attached to the prong of an electrically driven tuning fork which vibrates perpendicular to the length of the string at a frequency of 60 Hz. The weight (not mass) of the string is 0.096 lb. a) [5 pts] What tension must the string be under (weights are attached to the other end) if it is to vibrate in four loops

Answer 1

The tension in string will be "3.62 N".

Step-by-step explanation:

The given values are:

Length of string:

l = 3 ft

or,

= 0.9144 m

frequency,

f = 60 Hz

Weight,

= 0.096 lb

or,

= 0.0435 kgm/s²

Now,

The mass will be:

=
`(0.0435)/(9.8)`

=
`0.0044 \ kg`

As we know,

⇒
`\lambda=(2L)/(n)`

On substituting the values, we get

⇒
`=(2* 0.9144)/(4)`

⇒
`=0.4572 \ m`

or,

⇒
`v=f \lambda`

⇒
`=0.4572* 60`

⇒
`=27.432 \ m/s`

Now,

⇒
`v=\sqrt{(T)/(\mu) }`

or,

⇒
`T=(m)/(l)* v^2`

On putting the above given values, we get

⇒
`=(0.0044)/(0.9144)* (27.432)^2`

⇒
`=(752.51* 0.0044)/(0.9144)`

⇒
`=3.62 \ N`