Question

Question #b0f77

Answer 1

`rho_(Cu)=7575"kg/m"^3`

or `7.57"g/cm"^3`

Explanation:

The key to this question is to find the volume of the copper block.

Once we get this we can find the density since

density = mass / volume and the mass is given.

The relationship between the frequency `f` of the note and the tension `T` in the wire is given by:

`f=(1)/(2L)sqrt((T)/(mu))`

`L` is the length of the wire

`mu` is the mass per unit length of the wire

We will assume that these remain constant.

Diagram (a)

Since the wire is in tune with the fork we can say:

`f_1=96"Hz"`

If the block is immersed in water then Archimedes tells us that it will experience an upthrust which is equal to the weight of water displaced.

(freedive-earth.com)

This means that the tension in the wire will be slightly reduced which will result in the frequency being reduced.

(Think what happens to the note of a guitar string if it is loosened).

The 2 notes sounded together make "beats". The beat frequency is given by :

`f_b=Deltaf`

Where `Deltaf` is the magnitude of the difference between the 2 frequencies. Since we know that the new frequency must be lower we can say:

`f_2=96-5.6=90.4"Hz"`

So now we set up 2 equations to eliminate the constants:

`f_1=(1)/(2L)sqrt((T_1)/(mu))" "color(red)((1))`

and

`f_2=(1)/(2L)sqrt((T_2)/(mu))" "color(red)((2))`

Divide `color(red)((1))` by `color(red)((2))rArr`

`(f_1)/(f_2)=(cancel((1)/(2L))sqrt((T_1)/(cancel(mu))) )/(cancel((1)/(2L))sqrt((T_2)/(cancel(mu))))`

This simplifies down to:

`(f_1)/(f_2)=sqrt((T_1)/(T_2))`

Putting in the numbers:

`96/90.4=sqrt((10xx9.8)/(T_2))=1.0619`

`:. 1.0619 = sqrt(98/T_2`

`:. 1.0619^2=98/T_2=1.1277`

`:.T_2=98/1.1277=86.9"N"`

Diagram (b)

You can see from diagram (b) that from Archimedes the upthrust `U` provided by the water is given by:

`U=98-86.9=11.1"N"`

This equals the weight of water displaced so:

`mg=11.1`

`:.m=11.1/9.8=1.132"kg"`

We are given the density `rho` of the water so:

`rho=m/v`

`:.v=m/rho=1.132/1000=0.00132"m"^3`

This must equal the volume of the copper block.

`:.rho_(Cu)=m/v=10/0.00132=7575.7"kg/m"^3`