Question

The tension of the piano wire is $16kg-weight$. What must be the change in tension so as to produce a tone an octave lower?

• A. 4 kg weight
• B. 12 kg weight
• C. 32 kg weight
• D. 8 kg weight.

Answer 1

The correct option is B 12 kg weight

Given,

$T_1=16kg-weight$
Octave lower means, doubling the frequency($\eta$):

${\eta }_1:{\eta }_2=2:1$

The frequency in a stretched string is given by:

$\eta =\frac{1}{2l}\sqrt{\frac{T}{m}}$. . . . . . . .$\left(1\right)$

where, $l=$ Length of the wire

$T=$ Tension in the wire

$m=$ Mass of wire

From equation $\left(1\right)$,

$\eta \propto \sqrt{T}$

$\frac{{\eta }_1}{{\eta }_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{1}{2}=\sqrt{\frac{T_1}{T_2}}$

$T_2=\frac{T_1}{4}=\frac{16}{4}=4kg-weight$

Change in tension,

$\Delta T=T_1-T_2$

$\Delta T=16-4=12kg-weight$

The correct option is $\left(B\right)$.