🔊 بنك أسئلة الصوت

📌 Step 1: Speed of sound in air depends on temperature: \(v\propto\sqrt{T}\)

📌 Step 2: When temperature increases, air molecules move faster, so sound speed increases.

📌 Step 3: Direct relationship: as temperature increases, sound speed increases.

📌 Step 1: Sound speed depends on the medium's properties (elasticity and density).

📌 Step 2: In solids, molecules are tightly packed, making vibration transmission easier.

📌 Step 3: Speed order (fastest to slowest): solids > liquids > gases. Sound does not travel in vacuum.

📌 Step 1: Pitch is the property that makes a sound high or low.

📌 Step 2: Frequency is the number of vibrations per second, measured in Hertz (Hz).

📌 Step 3: Higher frequency → higher pitch. Amplitude determines loudness, not pitch.

📌 Step 1: Loudness (intensity) is how loud or soft a sound is perceived.

📌 Step 2: Amplitude is the maximum displacement from equilibrium. Larger amplitude = more energy.

📌 Step 3: Direct relationship: larger amplitude = louder sound.

📌 Step 1: The Doppler effect is a change in the apparent frequency when source and observer are in relative motion.

📌 Step 2: When the source approaches the observer, the apparent frequency increases (sound seems higher).

📌 Step 3: When the source moves away, the apparent frequency decreases (sound seems lower).

📌 Step 1: Standing waves in an air column form when sound reflects at the tube ends.

📌 Step 2: The main condition: the source frequency must equal the tube's natural resonance frequency (set by the tube's length and end conditions).

📌 Step 3: At resonance, the incident and reflected waves interfere to form a standing wave.

📌 Step 1: An open-open tube has antinodes at both ends (maximum displacement).

📌 Step 2: The fundamental frequency is the lowest resonance, with half a wavelength inside the tube.

📌 Step 3: Number of nodes = number of antinodes − 1. A node is a point of zero displacement.

📌 Step 1: Fundamental frequency for a closed-end tube: \(f_1=\dfrac{v}{4L}\)

📌 Step 2: Substitute: \(v=340\;m/s\), \(L=1\;m\)

📌 Step 3: \(f_1=\dfrac{340}{4\times1}=\dfrac{340}{4}=85\;Hz\)

📌 Step 1: For an open-open tube, the fundamental corresponds to \(\lambda=2L\)

📌 Step 2: Substitute the tube length: \(L=0.5\;m\)

📌 Step 3: \(\lambda=2\times0.5=1\;m\)

📌 Step 1: Sound level in decibels: \(\beta=10\log_{10}\left(\dfrac{I}{I_0}\right)\)

📌 Step 2: Difference in sound level: \(\Delta\beta=\beta_2-\beta_1=70-10=60\;dB\)

📌 Step 3: Relationship between intensity and pressure: \(I\propto P^2\)

📌 Step 4: \(\dfrac{I_2}{I_1}=10^{\Delta\beta/10}=10^6\), and since \(I\propto P^2\), \(\dfrac{P_2}{P_1}=\sqrt{10^6}=1000\)

📌 Step 1: General Doppler formula: \(f_d=f_s\cdot\dfrac{v+v_o}{v+v_s}\), where \(v_o\) is positive if the observer approaches the source, and \(v_s\) is positive if the source recedes from the observer.

📌 Step 2: The source (ambulance) is faster than the observer and moves in the same direction, so it is net receding from the observer: \(v_s=+40\;m/s\).

📌 Step 3: The observer moves toward the source's side (chasing it from behind), so it is approaching along the line: \(v_o=+20\;m/s\).

📌 Step 4: \(f_d=250\times\dfrac{344+20}{344+40}=250\times\dfrac{364}{384}\approx237\;Hz\)

📌 Note: A common mistake is treating the observer as 'receding' (since it's slower than the source) and using \(v_o=-20\); but the observer is actually moving toward the source's side (closing the gap at 20 m/s, only outpaced by the source's own motion), so the sign is positive. A direct wavefront-counting derivation confirms \(f_d\approx237\;Hz\), consistent with the next question.

📌 Step 1: Doppler effect: the more the source and observer approach each other, the higher the observed frequency.

📌 Step 2: The red car (C) and the ambulance are approaching each other (both heading toward one another) — the strongest frequency increase: highest pitch.

📌 Step 3: The man (D) is stationary, but the ambulance approaches him, so he hears a raised frequency (lower than C, since C also adds the observer's own approach).

📌 Step 4: The ambulance driver (A) has no relative motion with respect to the source, so hears the original frequency unchanged.

📌 Step 5: The black car (B) is behind the ambulance, moving slower in the same direction, so the ambulance is net receding from it — lowest observed frequency.

📌 Step 1: A closed-end tube has only odd-harmonic resonances: \(f_n=(2n-1)f_1\)

📌 Step 2: Since \(f_1=200\;Hz\), allowed frequencies are: \(200,\;600,\;1000,\;1400,\ldots\)

📌 Step 3: \(400\;Hz\) is not an odd multiple of \(200\;Hz\) (it's an even multiple), so it cannot be a resonance.

📌 Step 1: Closed tube: \(f_{closed}=\dfrac{v}{4L}\); open tube (same length L): \(f_{open}=\dfrac{v}{2L}\)

📌 Step 2: Relationship: \(f_{open}=2\times f_{closed}\)

📌 Step 3: \(f_{open}=2\times210=420\;Hz\)

📌 Step 1: Heating the air increases the speed of sound: \(v\propto\sqrt{T}\)

📌 Step 2: The frequency stays constant because it is set by the tuning fork (the source), not by the medium.

📌 Step 3: Since \(v=f\cdot\lambda\) and f is constant, increasing v increases \(\lambda\).

📌 Step 1: An open-open tube with 3 nodes and 4 antinodes means 3 half-wavelengths fit inside the tube (third harmonic).

📌 Step 2: Tube length \(L=3\times\dfrac{\lambda}{2}\), so \(\lambda=\dfrac{2L}{3}=\dfrac{2\times0.6}{3}=0.4\;m\)

📌 Step 3: Frequency \(f=\dfrac{v}{\lambda}=\dfrac{340}{0.4}=850\;Hz\)

📌 Step 4: Fundamental \(f_1=\dfrac{v}{2L}=\dfrac{340}{2\times0.6}=\dfrac{340}{1.2}\approx283.33\;Hz\) (check: \(f=3\times f_1\approx850\;Hz\) ✓)

📌 Step 1: A closed-end tube always has an EQUAL number of nodes and antinodes for each harmonic (unlike an open-open tube): 3 nodes and 3 antinodes correspond to harmonic \(2n-1=5\) (i.e. n=3).

📌 Step 2: Closed-tube condition: \(L=(2n-1)\times\dfrac{\lambda}{4}\), and here \(2n-1=5\), so \(L=5\times\dfrac{\lambda}{4}\)

📌 Step 3: \(\lambda=\dfrac{4L}{5}=\dfrac{4\times0.5}{5}=\dfrac{2}{5}=0.4\;m\)

📌 Step 4: Frequency \(f=\dfrac{v}{\lambda}=\dfrac{340}{0.4}=850\;Hz\)

📌 Step 5: Fundamental \(f_1=\dfrac{v}{4L}=\dfrac{340}{4\times0.5}=\dfrac{340}{2}=170\;Hz\) (check: \(f=5\times f_1=850\;Hz\) ✓)