Grade 11 : Physics - Waves (Set 2)

Explanation:Wave speed $v$ is related to frequency $f$ and wavelength $\lambda$ by the wave equation: $$v = f \lambda$$ Substituting the given values: $$v = 440\text{ Hz} \times 0.78\text{ m} = 343.2\text{ m/s}$$ Thus, Option A is correct.

Explanation:The period $T$ is the reciprocal of the frequency $f$: $$T = \frac{1}{f} = \frac{1}{250\text{ Hz}} = 0.004\text{ s}$$ Hence, Option B is correct.

Explanation:The distance between consecutive crests is the wavelength $\lambda$. Using $v = f \lambda$: $$\lambda = \frac{v}{f} = \frac{3.6\text{ m/s}}{3.0\text{ Hz}} = 1.20\text{ m}$$ Option B is correct.

Explanation:Phase difference $\Delta \phi$ is related to path difference $\Delta x$ by: $$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$$ Substituting $\lambda = 1.6m$ and $\Delta x = 0.4m$: $$\Delta \phi = \frac{2\pi}{1.6} \times 0.4 = \frac{\pi}{2}\text{ rad}$$ Option B is correct.

Explanation:The speed $v$ of a transverse wave on a stretched string is: $$v = \sqrt{\frac{T}{\mu}}$$ Given $T = 144\text{ N}$ and $\mu = 0.01\text{ kg/m}$: $$v = \sqrt{\frac{144}{0.01}} = \sqrt{14400} = 120\text{ m/s}$$ Option B is correct.

Explanation:According to the Doppler effect, when a wave source moves toward a stationary observer, the observed wave crests arrive at shorter time intervals, resulting in a higher apparent frequency. Option C is correct.

Explanation:By definition, in transverse waves medium particles oscillate perpendicular to the direction of wave propagation, whereas in longitudinal waves medium particles oscillate parallel to wave propagation. Option B is correct.

Explanation:Constructive interference occurs when path difference is an integer multiple of wavelength: $$\Delta x = n \lambda, \quad n = 0, 1, 2, \dots$$ For $n = 2$: $$\Delta x = 2 \times 0.5\text{ m} = 1.00\text{ m}$$ Option C is correct.

Explanation:For a string fixed at both ends, fundamental wavelength $\lambda_1 = 2L = 2 \times 1.2m = 2.4m$. Fundamental frequency is: $$f_1 = \frac{v}{2L} = \frac{240}{2.4} = 100\text{ Hz}$$ Option B is correct.

Explanation:For an open organ pipe, harmonic frequencies are $f_n = n \frac{v}{2L}$. For the second harmonic ($n = 2$): $$f_2 = \frac{v}{L} = \frac{340\text{ m/s}}{0.85\text{ m}} = 400\text{ Hz}$$ Option C is correct.

Explanation:For a pipe closed at one end, fundamental wavelength is $\lambda_1 = 4L$. Fundamental frequency: $$f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.425} = \frac{340}{1.7} = 200\text{ Hz}$$ Option B is correct.

Explanation:Beat frequency $f_{\text{beat}}$ is the absolute difference between two interfering frequencies: $$f_{\text{beat}} = |f_1 - f_2| = |444\text{ Hz} - 440\text{ Hz}| = 4\text{ Hz}$$ Option B is correct.

Explanation:The intensity $I$ of a wave is directly proportional to the square of its amplitude $A$ ($I \propto A^2$). Tripling amplitude ($A \to 3A$) results in: $$I' \propto (3A)^2 = 9A^2 = 9I$$ Option C is correct.

Explanation:Comparing with standard equation $y = A \sin(\omega t - kx)$, angular frequency $\omega = 40\pi\text{ rad/s}$ and wavenumber $k = 4\pi\text{ rad/m}$. Wave velocity is: $$v = \frac{\omega}{k} = \frac{40\pi}{4\pi} = 10\text{ m/s}$$ Option B is correct.

Explanation:Wavelength $\lambda = 12m$ and period $T = 4.0\text{ s}$. Wave speed is calculated as: $$v = \frac{\lambda}{T} = \frac{12\text{ m}}{4.0\text{ s}} = 3.0\text{ m/s}$$ Option B is correct.

Explanation:Total round-trip distance traveled by the sound wave is $2d = 2 \times 0.075m = 0.15m$. Time taken: $$t = \frac{2d}{v} = \frac{0.15\text{ m}}{1500\text{ m/s}} = 0.0001\text{ s} = 0.10\text{ ms}$$ Option B is correct.

Explanation:Average power transported by a harmonic wave is given by $P = \frac{1}{2} \mu v \omega^2 A^2 = 2\pi^2 \mu v f^2 A^2$. Thus, power depends directly on $f^2$ and $A^2$. Option B is correct.

Explanation:The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma R T}{M}}$. Therefore, $v$ is proportional to the square root of absolute temperature ($v \propto \sqrt{T}$). Option C is correct.

Explanation:Peak-to-trough vertical distance represents twice the amplitude ($2A = 0.60m$). Amplitude $A$ is: $$A = \frac{0.60\text{ m}}{2} = 0.30\text{ m}$$ Option B is correct.

Explanation:In any standing wave, the distance between two consecutive nodes is $\lambda/2$, and the distance between a node and the adjacent antinode is $\lambda/4$: $$d = \frac{1.6\text{ m}}{4} = 0.4\text{ m}$$ Option B is correct.