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

Answer:

The tension in string will be "3.62 N".

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:

= $\frac{0.0435}{9.8}$

= $0.0044 \ kg$

As we know,

⇒  $\lambda=\frac{2L}{n}$

On substituting the values, we get

⇒     $=\frac{2\times 0.9144}{4}$

⇒     $=0.4572 \ m$

or,

⇒  $v=f \lambda$

⇒      $=0.4572\times 60$

⇒      $=27.432 \ m/s$

Now,

⇒  $v=\sqrt{\frac{T}{\mu} }$

or,

⇒  $T=\frac{m}{l}\times v^2$

On putting the above given values, we get

⇒      $=\frac{0.0044}{0.9144}\times (27.432)^2$

⇒      $=\frac{752.51\times 0.0044}{0.9144}$

⇒      $=3.62 \ N$