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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?

2 Answers

6 votes

Answer:

Step-by-step 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=(1)/(2L)\sqrt{(T)/(\mu )}

where, μ = m/L (mass per unit length)

So,
m=(T)/(4Lf^(2))

By substituting the values


m=(667)/(4* 0.389* 440* 440)

m = 2.21 x 10^-3 kg

m = 2.21 g

User Tempire
by
3.5k points
3 votes

Answer:

Mass, m = 2.2 kg

Step-by-step 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=(1)/(2L)\sqrt{(T)/(\mu)}


\mu=(m)/(L) is the linear mass density

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


m=(T)/(4Lf^2)


m=(667)/(4* 0.389* (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.

User Bolzano
by
3.2k points