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

Respuesta :

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

[tex]W= mg[/tex]

[tex]m = \dfrac{W}{g}[/tex]

Put the value into the formula

[tex]m=\dfrac{81.34}{9.8}[/tex]

[tex]m=8.3\ kg[/tex]

The mass per unit length of the wire

[tex]m_{l}=\dfrac{m}{l}[/tex]

Put the value into the formula

[tex]m_{l}=\dfrac{8.3}{10000}[/tex]

[tex]m_{l}=8.3\times10^{-4}\ kg/m[/tex]

We need to calculate the frequency in the wire

Using formula of frequency

[tex]f_{w}=\dfrac{1}{2l}\sqrt{\dfrac{T}{m_{l}}}[/tex]

Put the value into the formula

[tex]f_{w}=\dfrac{1}{2\times0.660 }\sqrt{\dfrac{52}{8.3\times10^{-4}}}[/tex]

[tex]f_{w}=189.62\ Hz[/tex]

We need to calculate the beat frequency

Using formula of beat frequency

[tex]f_{b}=f_{in}-f_{w}[/tex]

Put the value into the formula

[tex]f_{b}=196-189.622[/tex]

[tex]f_{b}=6.378\ Hz[/tex]

Hence, The beat frequency is 6.378 Hz.