Question

From the given displacement – distance graph of a sound wave, find the value of $\text{AC – BC + CE - EF}$.​

• A. $5.75\text{cm}$
• B. $5.5\text{cm}$
• C. $4.25\text{cm}$
• D. $4.5\text{cm}$

Answer 1

The correct option is C $4.25\text{cm}$

• We know that the distance between any two consecutive crests or troughs in the wave represents the wavelength of a sound wave.
• Here $\text{BC}$ represents half of the wavelength. That means wavelength of the curve(𝛌) $=2\times 1.5=3\text{cm}$

$\text{AC = AB + BC}=𝛌+\frac{𝛌}{2}=\frac{3𝛌}{2}$
$\text{CE}=\frac{𝛌}{2}+\frac{𝛌}{4}=\frac{3𝛌}{4}$
$\text{EF}=1\text{cm}$ (amplitude)
Coming to $\text{AC – BC + CE - EF = AB + BC - BC + CE - EF}$
$\text{= AB + CE -EF}=𝛌+\frac{3𝛌}{4}-1=\frac{7𝛌}{4}-1$
Now put $𝛌=3$,
$\frac{7𝛌}{4}=\frac{7\times 3}{4}-1=5.25-1=4.25\text{cm}$