Class 11 - Unit # 12 : Acoustics - Solved Numericals

Physics Numerical Sheet
Acoustics

1. The equation of a wave is \(y(x, t) = 3.5 \sin \left\{ \frac{\pi}{3.0}x - 66t \right\}\text{ cm}\) where \(t\) is in seconds and \(x\) and \(y\) both are in cm. Find (a) the amplitude and (b) the wavelength of this wave.

Data:

• Given wave equation: \(y(x, t) = 3.5 \sin \left( \frac{\pi}{3.0}x - 66t \right)\)
• Amplitude (\(A\)) = ?
• Wavelength (\(\lambda\)) = ?

Solution:

The standard equation of a progressive sinusoidal wave moving along the positive x-axis is written as:

$$y(x, t) = A \sin (kx - \omega t) \quad \text{--- (ii)}$$

Comparing our given equation with the standard configuration:

$$\text{(a) Amplitude: } A = 3.5\text{ cm}$$

For the wave number (\(k\)):

$$k = \frac{\pi}{3.0}\text{ cm}^{-1}$$

Since we know that the wave number is related to the wavelength by \(k = \frac{2\pi}{\lambda}\):

$$\frac{2\pi}{\lambda} = \frac{\pi}{3.0} \implies \lambda = 2 \times 3.0 = 6.0\text{ cm}$$

RESULT: The amplitude of the wave is 3.5 cm and the wavelength is 6.0 cm.

2. Why is it that your own voice sounds strange to you when you hear it played back on a audio recorder, but your friends all agree that it is just what your voice sounds like?

Answer:

When you speak, your vocal cords create sound waves that travel through the air to reach your inner ear (**air conduction**). However, the bones, cartilage, and soft tissues in your skull also conduct those sound vibrations directly to your cochlea (**bone conduction**). The voice you hear inside your head while speaking is a blend of both paths.

Bone conduction filters and enhances **lower-frequency vibrations**, making your voice sound fuller, deeper, and richer to yourself. When you listen to a recording, the bone-conduction path is absent, and you only receive the sound via air conduction. Because you miss the deep, resonant frequencies your skull usually provides, your voice sounds higher-pitched, thinner, and unfamiliar to you—even though it is exactly what everyone else hears.

3. Why is the speed of sound in solids much faster than the speed of sound in air?

Answer:

Sound is a mechanical longitudinal wave that propagates through consecutive collisions between adjacent particles. The velocity of these waves depends heavily on the **elasticity** and **density** of the medium (\(v = \sqrt{E/\rho}\)).

Even though solids are denser than gases, they have an incredibly high elastic modulus (they resist deformation strongly and snap back into place extremely quickly). The tightly packed atomic arrangement in a solid enables mechanical disturbances to pass to neighboring atoms with minimal delay. In contrast, gas molecules are far apart and must drift across a space before hitting a neighbor, creating a significant delay that drastically slows down the speed of sound in air.

4. An increase in pressure of 100 kPa causes a certain volume of water to decrease by \(5 \times 10^{-3}\%\) of its original volume. Find (a) the Bulk modulus of water and (b) the speed of sound in water.

Data:

• Increase in pressure (\(\Delta P\)) = \(100\text{ kPa} = 100 \times 10^3\text{ Pa} = 1 \times 10^5\text{ N/m}^2\)
• Volumetric strain percentage = \(5 \times 10^{-3}\%\)
• Fractional change in volume (\(\frac{\Delta V}{V}\)) = \(\frac{5 \times 10^{-3}}{100} = 5 \times 10^{-5}\)
• Density of water (\(\rho\)) = \(1000\text{ kg/m}^3\)
• Bulk Modulus (\(B\)) = ?
• Speed of sound (\(v\)) = ?

Solution:

Part (a): According to the definition of Bulk Modulus:

$$B = \frac{\Delta P}{\frac{\Delta V}{V}}$$ $$B = \frac{1 \times 10^5}{5 \times 10^{-5}} = 2.0 \times 10^9\text{ N/m}^2$$

Part (b): The speed of sound through a fluid medium is calculated using the Newton-Laplace formula:

$$v = \sqrt{\frac{B}{\rho}}$$ $$v = \sqrt{\frac{2.0 \times 10^9}{1000}} = \sqrt{2.0 \times 10^6} \approx 1414.2\text{ m/s}$$

RESULT: The bulk modulus of water is \(2.0 \times 10^9\text{ N/m}^2\) and the speed of sound in water is \(1414.2\text{ m/s}\).

5. A uniform string of length 10.0 m and weight 0.25 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. The lower end of the string is suddenly displaced horizontally. How long does it take the resulting wave pulse to travel to the upper end? (Neglect the weight of the string in comparison to the hanging mass).

Data:

• Length of string (\(L\)) = \(10.0\text{ m}\)
• Weight of string (\(W_s\)) = \(0.25\text{ N}\)
• Hanging load / Tension (\(T\)) = \(1.00\text{ kN} = 1000\text{ N}\)
• Acceleration due to gravity (\(g\)) = \(9.8\text{ m/s}^2\)
• Travel time (\(t\)) = ?

Solution:

The velocity of a transverse wave pulse travelling on a stretched string depends on the tension and the linear mass density:

$$v = \sqrt{\frac{T}{\mu}} \quad \text{--- (i)}$$

First, let's find the mass per unit length (\(\mu\)) using the string's mass (\(m = \frac{W_s}{g}\)):

$$m = \frac{0.25}{9.8} \approx 0.0255\text{ kg}$$ $$\mu = \frac{m}{L} = \frac{0.0255}{10.0} = 0.00255\text{ kg/m}$$

Substituting \(\mu\) back into the velocity equation (i):

$$v = \sqrt{\frac{1000}{0.00255}} = \sqrt{392156.8} \approx 626.22\text{ m/s}$$

The time required for the pulse to clear the entire distance is:

$$t = \frac{L}{v} = \frac{10.0}{626.22} \approx 0.01597\text{ s} \approx 16\text{ ms}$$

RESULT: It takes approximately 16 ms (\(0.016\text{ s}\)) for the wave pulse to travel to the upper end.

6. A travelling sine wave is the result of the superposition of two other sine waves with equal amplitudes, wavelengths, and frequencies. The two component waves each have an amplitude of 5.00 cm. If the resultant wave has an amplitude of 6.69 cm, what is the phase difference \(\phi\) between the component waves?

Data:

• Component wave amplitudes (\(A_1 = A_2 = A\)) = \(5.00\text{ cm}\)
• Resultant wave amplitude (\(A_R\)) = \(6.69\text{ cm}\)
• Phase difference (\(\phi\)) = ?

Solution:

From the principle of superposition, the amplitude of a resultant wave generated by two interfering waves of identical amplitude is given by:

$$A_R = 2A \cos\left(\frac{\phi}{2}\right)$$

Isolating the phase angle term:

$$6.69 = 2(5.00) \cos\left(\frac{\phi}{2}\right)$$ $$6.69 = 10.0 \cos\left(\frac{\phi}{2}\right) \implies \cos\left(\frac{\phi}{2}\right) = 0.669$$

Taking the inverse cosine:

$$\frac{\phi}{2} = \cos^{-1}(0.669) \approx 48^{\circ}$$ $$\phi = 48^{\circ} \times 2 = 96^{\circ} \quad \text{(or } 1.68\text{ radians)}$$

RESULT: The phase difference \(\phi\) between the component waves is \(96^{\circ}\) (or \(1.68\text{ rad}\)).

7. In order to decrease the fundamental frequency of a guitar string by 4.0%, by what percentage should you reduce the tension?

Data:

• Initial Frequency = \(f_1\)
• Target Frequency (\(f_2\)) = \(f_1 - 4\% \text{ of } f_1 = 0.96 f_1\)
• Percentage reduction in tension (\(\frac{\Delta T}{T_1} \times 100\)) = ?

Solution:

The fundamental frequency of a stretched string fixed at both ends is modeled as:

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

Setting up the ratio for the initial and final states:

$$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} \quad \text{--- (i)}$$

Substitute the frequency criteria \(f_2 = 0.96 f_1\) into the expression:

$$\frac{0.96 f_1}{f_1} = \sqrt{\frac{T_2}{T_1}} \implies 0.96 = \sqrt{\frac{T_2}{T_1}}$$

Squaring both sides to find the remaining tension fraction:

$$\frac{T_2}{T_1} = (0.96)^2 = 0.9216$$ $$T_2 = 0.9216 T_1$$

The required reduction in tension is calculated as:

$$\text{Percentage Reduction} = \left(1 - \frac{T_2}{T_1}\right) \times 100$$ $$\text{Percentage Reduction} = (1 - 0.9216) \times 100 = 7.84\%$$

RESULT: The tension must be reduced by 7.84%.

8. A string 2.0 m long is held fixed at both ends. If a sharp blow is applied to the string at its centre, it takes 0.050 s for the pulse to travel to both ends of the string and return to the middle. What is the fundamental frequency of oscillation for this string?

Data:

• Total length of string (\(L\)) = \(2.0\text{ m}\)
• Time elapsed (\(t\)) = \(0.050\text{ s}\)
• Total round-trip distance traveled by a pulse (\(d\)) = \(\frac{L}{2} \text{ (to end)} + \frac{L}{2} \text{ (back)} = L = 2.0\text{ m}\)
• Fundamental frequency (\(f_1\)) = ?

Solution:

First, we calculate the propagation velocity (\(v\)) of the wave pulse along the string framework:

$$v = \frac{\text{Distance}}{\text{Time}} = \frac{2.0\text{ m}}{0.050\text{ s}} = 40\text{ m/s}$$

The fundamental frequency for a string clamped tightly at both boundaries is defined by:

$$f_1 = \frac{v}{2L}$$ $$f_1 = \frac{40}{2 \times 2.0} = \frac{40}{4.0} = 10\text{ Hz}$$

RESULT: The fundamental frequency of oscillation is 10 Hz.

10. A train sounds its whistle while passing by a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches the crossing and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Find the speed of the train and the frequency of its whistle.

Data:

• Apparent approach frequency (\(f'\)) = \(219\text{ Hz}\)
• Apparent recede frequency (\(f''\)) = \(184\text{ Hz}\)
• Speed of sound (\(v\)) = \(340\text{ m/s}\)
• Speed of the source/train (\(v_s\)) = ?
• Actual frequency of the whistle (\(f\)) = ?

Solution:

Using the classical Doppler shift formulations for a moving source and stationary listener:

$$\text{Case 1 (Approaching): } f' = f \left( \frac{v}{v - v_s} \right) \quad \text{--- (i)}$$ $$\text{Case 2 (Receding): } f'' = f \left( \frac{v}{v + v_s} \right) \quad \text{--- (ii)}$$

Dividing equation (i) by equation (ii) to eliminate the true frequency variable (\(f\)):

$$\frac{f'}{f''} = \frac{v + v_s}{v - v_s} \implies \frac{219}{184} = \frac{340 + v_s}{340 - v_s}$$ $$1.1902 = \frac{340 + v_s}{340 - v_s}$$

Cross-multiplying and solving for \(v_s\):

$$1.1902(340 - v_s) = 340 + v_s$$ $$404.67 - 1.1902v_s = 340 + v_s$$ $$64.67 = 2.1902v_s \implies v_s = \frac{64.67}{2.1902} \approx 29.53\text{ m/s}$$

Now, calculate the true whistle frequency by substituting \(v_s\) back into equation (i):

$$219 = f \left( \frac{340}{340 - 29.53} \right)$$ $$219 = f \left( \frac{340}{310.47} \right) \implies 219 = f(1.0951)$$ $$f = \frac{219}{1.0951} \approx 200\text{ Hz}$$

RESULT: The speed of the train is 29.53 m/s and the actual frequency of its whistle is 200 Hz.