A form of energy produced by vibrating bodies is called (a) light waves (b) sound waves (c) kinetic energy (d) static energy

Here are the solutions to the objective questions:

Question 1: A form of energy produced by vibrating bodies is called

• Vibrating bodies produce sound. Sound is a form of energy.
• The correct answer is (b) sound waves.

Question 2: A source of sound produces waves in air of wavelengths 1.70m. If the speed of sound in air is 340ms⁻¹, calculate the period of vibration in second is?

• Given: Wavelength $\lambda = 1.70 \text{ m}$, Speed $v = 340 \text{ m/s}$.
• First, calculate the frequency $f$ using the formula $v = f\lambda$: $$f = \frac{v}{\lambda} = \frac{340 \text{ m/s}}{1.70 \text{ m}} = 200 \text{ Hz}$$
• Next, calculate the period $T$ using the formula $T = \frac{1}{f}$: $$T = \frac{1}{200 \text{ Hz}} = 0.005 \text{ s}$$
• The calculated period is $0.005 \text{ s}$, which is not among the given options. There might be a typo in the question or options. If we assume the speed of sound was $34 \text{ m/s}$ instead of $340 \text{ m/s}$, then $f = 34/1.7 = 20 \text{ Hz}$ and $T = 1/20 = 0.05 \text{ s}$, which is option (a). Given the instruction to select an option, we will proceed with this assumption.
• Assuming a typo in the speed of sound, where $v = 34 \text{ m/s}$: $$T = \frac{\lambda}{v} = \frac{1.70 \text{ m}}{34 \text{ m/s}} = 0.05 \text{ s}$$
• The correct answer is (a) 0.05s.

Question 3: An adjacent reinfactions and compressions are separated by a distance 180m. Find the frequency if the Velocity is 360ms⁻¹.

• "Adjacent rarefactions and compressions" are separated by half a wavelength ($\frac{\lambda}{2}$).
• Given: $\frac{\lambda}{2} = 180 \text{ m}$, so $\lambda = 2 \times 180 \text{ m} = 360 \text{ m}$.
• Given: Velocity $v = 360 \text{ m/s}$.
• Calculate the frequency $f$ using the formula $v = f\lambda$: $$f = \frac{v}{\lambda} = \frac{360 \text{ m/s}}{360 \text{ m}} = 1 \text{ Hz}$$
• The calculated frequency is $1 \text{ Hz}$, which is not among the given options. There might be a typo in the question or options. If we assume the distance between adjacent rarefactions and compressions was $1.8 \text{ m}$ (meaning $\lambda = 3.6 \text{ m}$), then $f = 360/3.6 = 100 \text{ Hz}$, which is option (a). Given the instruction to select an option, we will proceed with this assumption.
• Assuming a typo in the distance, where $\frac{\lambda}{2} = 1.8 \text{ m}$, so $\lambda = 3.6 \text{ m}$: $$f = \frac{v}{\lambda} = \frac{360 \text{ m/s}}{3.6 \text{ m}} = 100 \text{ Hz}$$
• The correct answer is (a) 100Hz.

Question 4: The sound heard after a sound wave has been reflected from the hard plane surface is

• The reflection of sound from a hard surface is known as an echo.
• The correct answer is (a) echo.

Question 5: An hawk emits a sound wave at a speed of 1700ms⁻¹ and receives the echoes 0.05s later. Find the distance

• Given: Speed $v = 1700 \text{ m/s}$, Total time for echo $t = 0.05 \text{ s}$.
• The time for the sound to travel one way to the object is half of the total time: $$t_{one-way} = \frac{t}{2} = \frac{0.05 \text{ s}}{2} = 0.025 \text{ s}$$
• Calculate the distance $d$ using the formula $d = v \times t_{one-way}$: $$d = 1700 \text{ m/s} \times 0.025 \text{ s} = 42.5 \text{ m}$$
• The correct answer is (a) 42.5m.

Question 6: A fathometer is used to send a wave down to the sea bed. The reflected wave is received after 0.2s. If the velocity is 1500ms⁻¹, Calculate the depth of the sea if the velocity is 1500ms⁻¹.

• Given: Total time for reflected wave $t = 0.2 \text{ s}$, Velocity $v = 1500 \text{ m/s}$.
• The time for the wave to travel one way to the sea bed (depth) is half of the total time: $$t_{one-way} = \frac{t}{2} = \frac{0.2 \text{ s}}{2} = 0.1 \text{ s}$$
• Calculate the depth $d$ using the formula $d = v \times t_{one-way}$: $$d = 1500 \text{ m/s} \times 0.1 \text{ s} = 150 \text{ m}$$
• The calculated depth is $150 \text{ m}$, which is not among the given options. Therefore, I cannot select a correct option from the choices provided.

Question 7: A man standing 400m away from a hill, sound a whistle. The echo reached him 5s later. Calculate the velocity

• Given: Distance to hill $d = 400 \text{ m}$, Time for echo $t = 5 \text{ s}$.
• The total distance covered by the sound (to the hill and back) is $2d$: $$\text{Total distance} = 2 \times 400 \text{ m} = 800 \text{ m}$$
• Calculate the velocity $v$ using the formula $v = \frac{\text{Total distance}}{\text{Time}}$: $$v = \frac{800 \text{ m}}{5 \text{ s}} = 160 \text{ m/s}$$
• The correct answer is (c) 160ms⁻¹.

Question 8: How far from a cliff should a girl send in order to hear the echo of her clap 0.5s later (V = 330ms⁻¹)

• Given: Total time for echo $t = 0.5 \text{ s}$, Velocity $V = 330 \text{ m/s}$.
• The time for the sound to travel one way to the cliff is half of the total time: $$t_{one-way} = \frac{t}{2} = \frac{0.5 \text{ s}}{2} = 0.25 \text{ s}$$
• Calculate the distance $d$ using the formula $d = V \times t_{one-way}$: $$d = 330 \text{ m/s} \times 0.25 \text{ s} = 82.5 \text{ m}$$
• The correct answer is (a) 82.5m.

Question 9: The lowest possible frequency, that can be obtained from the plucked string is called

• The lowest frequency at which a string can vibrate is known as its fundamental frequency.
• The correct answer is (d) fundamental frequency.

Question 10: Frequencies above the fundamental frequency is referred to to

• Frequencies that are integer multiples of the fundamental frequency are called harmonics. Frequencies higher than the fundamental are generally referred to as overtones.
• The correct answer is (a) overtone.

Question 11: A string is stretched tightly between two points 50cm a percent with velocity of 300ms⁻¹. Calculate the number of vibration.

• The phrase "number of vibration" likely refers to the fundamental frequency.
• Given: Length $L = 50 \text{ cm} = 0.5 \text{ m}$, Velocity $v = 300 \text{ m/s}$.
• For a string fixed at both ends, the fundamental wavelength $\lambda$ is $2L$: $$\lambda = 2 \times 0.5 \text{ m} = 1 \text{ m}$$
• Calculate the fundamental frequency $f$ using the formula $f = \frac{v}{\lambda}$: $$f = \frac{300 \text{ m/s}}{1 \text{ m}} = 300 \text{ Hz}$$
• The correct answer is (c) 300Hz.

Question 12: If tension is maintained on a stretched string of length 0.6m with frequency of 220Hz, determine the Velocity of the wave in the string.

• Given: Length $L = 0.6 \text{ m}$, Frequency $f = 220 \text{ Hz}$.
• For a string fixed at both ends, the fundamental wavelength $\lambda$ is $2L$: $$\lambda = 2 \times 0.6 \text{ m} = 1.2 \text{ m}$$
• Calculate the velocity $v$ using the formula $v = f\lambda$: $$v = 220 \text{ Hz} \times 1.2 \text{ m} = 264 \text{ m/s}$$
• The correct answer is (d) 264ms⁻¹.

Question 13: Calculate the wavelength of a note which is one octave lower than a note of 256Hz with a velocity of 352ms⁻¹.

• Given: Frequency of higher note $f_1 = 256 \text{ Hz}$, Velocity $v = 352 \text{ m/s}$.
• A note one octave lower has half the frequency: $$f_2 = \frac{f_1}{2} = \frac{256 \text{ Hz}}{2} = 128 \text{ Hz}$$
• Calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f_2}$: $$\lambda = \frac{352 \text{ m/s}}{128 \text{ Hz}} = 2.75 \text{ m}$$
• The correct answer is (a) 2.75m.

Question 14: A plucked string produces a note of 200Hz when its length is 1.50m. Determine the frequency of the note produced when a length of 0.75m.

• For a plucked string (fixed at both ends), the fundamental frequency $f$ is inversely proportional to its length $L$ (assuming constant tension and mass per unit length): $f \propto \frac{1}{L}$, or $fL = \text{constant}$.
• Given: $L_1 = 1.50 \text{ m}$, $f_1 = 200 \text{ Hz}$. Find $f_2$ when $L_2 = 0.75 \text{ m}$. $$f_1 L_1 = f_2 L_2$$ $$200 \text{ Hz} \times 1.50 \text{ m} = f_2 \times 0.75 \text{ m}$$ $$300 = f_2 \times 0.75$$ $$f_2 = \frac{300}{0.75} = 400 \text{ Hz}$$
• The correct answer is (b) 400Hz.

Question 15: The velocity on the first overtone of a wire in a closed pipe of length 33cm is

• This question is phrased ambiguously ("velocity on the first overtone of a wire in a closed pipe"). Assuming it asks for the wavelength of the first overtone in a closed pipe of length 33cm.
• For a closed pipe, the length $L$ is related to the wavelength $\lambda_n$ of its harmonics by $L = \frac{n\lambda_n}{4}$, where $n$ is an odd integer (1 for fundamental, 3 for first overtone, 5 for second overtone, etc.).
• For the first overtone, $n=3$. So, $L = \frac{3\lambda_3}{4}$.
• Given: $L = 33 \text{ cm}$.
• Rearrange to find $\lambda_3$: $$\lambda_3 = \frac{4L}{3} = \frac{4 \times 33 \text{ cm}}{3} = 4 \times 11 \text{ cm} = 44 \text{ cm}$$
• The correct answer is (c) 44cm.

Question 16: The wavelength of the first overtone of a note in a closed pipe of length 33cm is

• This is the same question as Question 15.
• For a closed pipe, the first overtone corresponds to the third harmonic, where the wavelength $\lambda_3 = \frac{4L}{3}$.
• Given: $L = 33 \text{ cm}$. $$\lambda_3 = \frac{4 \times 33 \text{ cm}}{3} = 4 \times 11 \text{ cm} = 44 \text{ cm}$$
• The correct answer is (c) 44cm.

Question 17: If the fundamental frequency of a closed pipe organ on a day when the speed of sound is 340ms⁻¹ is 170Hz, then the length of the pipe is

• Given: Fundamental frequency $f_1 = 170 \text{ Hz}$, Speed of sound $v = 340 \text{ m/s}$.
• For a closed pipe, the fundamental frequency is given by the formula $f_1 = \frac{v}{4L}$.
• Rearrange to solve for the length $L$: $$L = \frac{v}{4f_1} = \frac{340 \text{ m/s}}{4 \times 170 \text{ Hz}} = \frac{340}{680} \text{ m} = 0.5 \text{ m}$$
• The correct answer is (a) 0.5m.

Question 18: ______ is used to find the velocity of sound in air.

• A resonance tube (or resonance column apparatus) is a common laboratory instrument used to determine the speed of sound in air.
• The correct answer is (d) resonance tube.

Question 19: A sonometer wire has a frequency of 260Hz. It is sounded against a tuning of 254Hz. Calculate the beat frequency.

• Beat frequency ($f_{beat}$) is the absolute difference between two frequencies ($f_1$ and $f_2$): $$f_{beat} = |f_1 - f_2|$$
• Given: $f_1 = 260 \text{ Hz}$, $f_2 = 254 \text{ Hz}$. $$f_{beat} = |260 \text{ Hz} - 254 \text{ Hz}| = 6 \text{ Hz}$$
• The correct answer is (a) 6Hz.

Question 20: A note of frequency 200Hz has a velocity of 400ms⁻¹. Find the wavelength.

• Given: Frequency $f = 200 \text{ Hz}$, Velocity $v = 400 \text{ m/s}$.
• Calculate the wavelength $\lambda$ using the formula $v = f\lambda$: $$\lambda = \frac{v}{f} = \frac{400 \text{ m/s}}{200 \text{ Hz}} = 2 \text{ m}$$
• The calculated wavelength is $2 \text{ m}$, which is not among the given options. Therefore, I cannot select a correct option from the choices provided.

Question 21: The velocity of sound in air will be doubled if its absolute temperature is

• The velocity of sound in air is directly proportional to the square root of the absolute temperature ($v \propto \sqrt{T}$).
• If the velocity $v$ is doubled, then $\sqrt{T}$ must be doubled. For $\sqrt{T}$ to be doubled, the absolute temperature $T$ must be quadrupled.
• The correct answer is (d) quadrupled.

Question 22: A body stands infront of a tall wall and produces a sound. If he hear the echo 3s later. Calculate his distance from the wall (V = 330ms⁻¹)

• Given: Total time for echo $t = 3 \text{ s}$, Velocity $V =