Question

A solid copper sphere hangs at the bottom of a steel wire of negligible mass. The top end of the wire is fixed. When the wire is struck, it vibrates with a fundamental frequency of 300 Hz. The copper object is then submerged in water so that half its volume is below the water line. Determine the new fundamental frequency. The density of water, copper, and steel is 1000kg/m³, 8960kg/m³, and 8050kg/m³ respectively

Answer 1

Answer:

291.509 Hz

Explanation:

Fundamental frequency, often referred to simply as the fundamental, is defined as the lowest frequency of a periodic waveform.

It is a vital concept in musical instruments and many aspects .

See the attached file for the solution to the given problem.

Answer 2

Answer:

Explanation:

The problem is based on vibration in string .For it the formula is as follows :

n = k √T , n is frequency of fundamental tone , k is constant for a string of given length and T is tension in the string .

In the first case weight of sphere is equal to tension .

weight = volume of sphere x density of copper x g

= v x 8960 x g

Putting the values

300 = k √ (v x 8960 x g)

300 = k √ 87808v

When the sphere is half submerged

tension is equal to apparent weight of sphere in water

=  v x 8960 x g - .5  v x 1000 x 9.8 , v is volume

= 82908v

for second case

n =  k √ 82908v

dividing the two equations

n / 300 = $\sqrt{\frac{82908}{87808} }$

$\frac{n}{300}$ = .9716

n = 291.5 Hz .