Question

A string is stretched to a length of 376 cm and
both ends are fixed.
If the density of the string is 0.036 g/cm,
and its tension is 574 N, what is the fundamental frequency?
Answer in units of Hz.

Answer 1

Answer: 53.09Hz

Explanation:

The fundamental frequency of an ideal taut string is:

Fn= n/2L(√T/μ)

Where:

F= frequency per second (Hz)

T= Tension of the string (cm/s sqr)

L= Length of the string (cm)

μ= Linear density or mass per unit length of the string in cm/gm

√T/μ= square root of T divided by μ

It is important to note:

Note: Typically, tension would be in newtons, length in meters and linear density in kg/m, but those units are inconvenient for calculations with strings. Here, the smaller units are used.

F1= 1/2(376cm)(0.01/1) × (√574/(0.036g/cm)(0.1kg/m÷1g/cm)

F1= 0.1329 × 399.30

= 53.09Hz

Answer 2

The fundamental frequency of the string is 51.97 Hz.

To calculate the fundamental frequency of the string, we use the formula below.

Formula:

• F' = (1/2l)√(T/m)............... Equation 1

Where:

• F' = Fundamental frequency of the string
• l = length of the string
• T = Tension on the string
• m = mass per unit length of the string

From the question,

Given:

• l = 376 cm = 3.76 m
• T = 574 N
• m = 0.036 g/cm = 0.0036 kg/m

Substitute these values into equation 1

• F' = 1/(2×3.76)[√(574/0.0036)]
• F' = (0.133){√(152659.57)
• F' = (0.133×390.72)
• F' = 51.97 Hz.

Hence, the fundamental frequency of the string is 51.97 Hz.

Learn more about fundamental frequency here: https://brainly.com/question/1967686