185k views
2 votes
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

User Robertwest
by
6.3k points

1 Answer

3 votes

Answer:

The beat frequency is 6.378 Hz.

Step-by-step 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 = (W)/(g)

Put the value into the formula


m=(81.34)/(9.8)


m=8.3\ kg

The mass per unit length of the wire


m_(l)=(m)/(l)

Put the value into the formula


m_(l)=(8.3)/(10000)


m_(l)=8.3*10^(-4)\ kg/m

We need to calculate the frequency in the wire

Using formula of frequency


f_(w)=(1)/(2l)\sqrt{(T)/(m_(l))}

Put the value into the formula


f_(w)=(1)/(2*0.660 )\sqrt{(52)/(8.3*10^(-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.

User Slaporte
by
6.0k points