Question

1. A huge spool of wire, 10,000 meters long, weighs 81.34 N. You cut off a meter or so and tie it between two posts, 0.660 m apart. The tension in the wire is set to 52 N. When the string is plucked, at the same instant a 196 Hz tuning fork is also hit, what beat frequency is heard

Answer 1

Answer:

The beat frequency is 6.378 Hz.

Explanation:

Given that,

Length of wire = 10000 m

Weight = 81.34 N

Distance = 0.660 m

Tension = 52 N

Frequency = 196 Hz

We need to calculate the mass of the wire

Using formula of weight

$W= mg$

$m = \dfrac{W}{g}$

Put the value into the formula

$m=\dfrac{81.34}{9.8}$

$m=8.3\ kg$

The mass per unit length of the wire

$m_{l}=\dfrac{m}{l}$

Put the value into the formula

$m_{l}=\dfrac{8.3}{10000}$

$m_{l}=8.3\times10^{-4}\ kg/m$

We need to calculate the frequency in the wire

Using formula of frequency

$f_{w}=\dfrac{1}{2l}\sqrt{\dfrac{T}{m_{l}}}$

Put the value into the formula

$f_{w}=\dfrac{1}{2\times0.660 }\sqrt{\dfrac{52}{8.3\times10^{-4}}}$

$f_{w}=189.62\ Hz$

We need to calculate the beat frequency

Using formula of beat frequency

$f_{b}=f_{in}-f_{w}$

Put the value into the formula

$f_{b}=196-189.622$

$f_{b}=6.378\ Hz$

Hence, The beat frequency is 6.378 Hz.