Question

A piano wire has a length of 81 cm and a mass of 2.0

g. if its fundamental frequency is to be 394 hz, its tension must be

Answer 1

To achieve a fundamental frequency of 394 Hz, a piano wire with a length of 81 cm and a mass of 2.0 g requires a tension of approximately 779.04 newtons.

Step-by-step explanation:

To calculate the tension necessary for a piano wire to have a fundamental frequency of 394 Hz, we can use the formula for the fundamental frequency of a stretched string, which is:

f = (1/2L) × √(T/μ), where:

• f is the fundamental frequency,
• L is the length of the wire,
• T is the tension in the wire, and
• μ is the linear mass density of the wire (mass per unit length).

First, we must calculate the linear mass density (μ = m/L) of the wire:

μ = 2.0 g / 81 cm = 0.02469 g/cm = 0.0002469 kg/m (after converting to SI units).

Now, we rearrange the formula to solve for T:

T = (2Lf)² × μ

T = (2 × 0.81 m × 394 Hz)² × 0.0002469 kg/m

T = 779.04 N approximately.

Therefore, the piano tuner must apply a force equivalent to a tension of about 779.04 newtons to make the wire resonate at a fundamental frequency of 394 Hz.

Answer 2

Frequency = 394 Hz
Length of the string L = 81 cm = 0.81 m
Mass of the string = 0.002 kg
Tension T = ?
Wave length of the string is two times the length.
n x lambda = 2L, we also have lambda = vt = v / f, t is time period and given n = 1.
Therefore L = v / 2f => v = 2fL
Deriving form force equation, force here is tension T so
v = squareroot of (TL/m) hence
2fL = squareroot of (TL/m) => 4 x f^2 x L^2 = (T x L) / m => T = 4 x f^2 x L x m
T = 4 x 0.81 x (394)^2 x 0.002 = 4 x 0.81 x 155236 x 0.002
T = 1005.9 N = 1.006 x 10^3 N