Question

A string of length 1.56 meters is used to hang a sphere of mass

5.00 kg. This system is at rest. If the mass of the string is 40.0
grams, what is the frequency of the 3rd harmonic wave on the
string?

Answer 1

To find the third harmonic frequency, the mass and length of the string allow us to calculate its linear density and the tension it experiences from the hanging sphere. These values are substituted into the formula for the frequency of the nth harmonic of a string fixed at both ends, yielding approximately 70.00 Hz for the third harmonic.

Step-by-step explanation:

To determine the frequency of the third harmonic of a wave on a string, we use the basic formula for the frequency of the nth harmonic of a string fixed at both ends:

f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}

Where:

• f_n is the frequency of the nth harmonic
• n is the harmonic number
• L is the length of the string
• T is the tension in the string
• \mu is the linear mass density of the string (mass per unit length)

To find the tension T, we use the weight of the sphere:

T = m \cdot g

Where m is the mass of the sphere and g is the acceleration due to gravity (approximately 9.81 m/s2). The linear mass density \mu of the string is the mass of the string divided by its length.

Now, applying the values:

\mu = \frac{0.040 kg}{1.56 m} = 0.02564 kg/m

T = 5.00 kg \cdot 9.81 m/s2 = 49.05 N

The frequency of the third harmonic (n=3) is:

f_3 = \frac{3}{2 \cdot 1.56 m} \sqrt{\frac{49.05 N}{0.02564 kg/m}}

f_3 = \frac{3}{3.12 m} \sqrt{\frac{49.05 N}{0.02564 kg/m}}

f_3 = \frac{3}{3.12} \cdot \sqrt{\frac{49.05}{0.02564}}

f_3 \approx 1.60 \cdot \sqrt{1912.50}

f_3 \approx 1.60 \cdot 43.73 Hz

f_3 \approx 70.00 Hz