Question

The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's its mass?

Answer 1

Answer:

Explanation:

frequency, f = 440 Hz

length, L = 38.9 cm = 0.389 m

Tension, T = 667 N

The formula for the frequency is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

where, μ = m/L (mass per unit length)

So, $m=\frac{T}{4Lf^{2}}$

By substituting the values

$m=\frac{667}{4\times 0.389\times 440\times 440}$

m = 2.21 x 10^-3 kg

m = 2.21 g

Answer 2

Answer:

Mass, m = 2.2 kg                                

Explanation:

It is given that,

Frequency of the piano, f = 440 Hz

Length of the piano, L = 38.9 cm = 0.389 m

Tension in the spring, T = 667 N

The frequency in the spring is given by :

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$

$\mu=\dfrac{m}{L}$ is the linear mass density

On rearranging, we get the value of m as follows :

$m=\dfrac{T}{4Lf^2}$

$m=\dfrac{667}{4\times 0.389\times (440)^2}$

m = 0.0022 kg

or

m = 2.2 grams

So, the mass of the object is 2.2 grams. Hence, this is the required solution.