A reflected sound returns after 0.6 s. If sound speed is 340 m/s, how far away is the wall?
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The sound travels to the wall and back, so distance to wall is vt/2.
340×0.6/2 = 102 m.
Practice ECAT questions with answers and explanations.
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The sound travels to the wall and back, so distance to wall is vt/2.
340×0.6/2 = 102 m.
Choose an option to check your answer.
A converging lens brings paraxial parallel rays to its principal focus.
The focal length is positive under the common sign convention.
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In-phase displacements add constructively.
The resultant amplitude is 3+2 = 5 cm.
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A diverging lens causes rays to separate.
Their backward extensions meet at a virtual focal point.
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Opposite-phase amplitudes subtract.
The resultant amplitude is |5-3| = 2 cm.
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For the fundamental, the string length is half a wavelength.
λ=2L=2.4 m.
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The fundamental wavelength is 2L=1.5 m.
F=v/λ=300/1.5 = 200 Hz.
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For an open pipe, fundamental wavelength is 2L=1.7 m.
F=340/1.7 = 200 Hz.
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For a closed pipe, fundamental wavelength is 4L=3.4 m.
F=340/3.4 = 100 Hz.
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Reflection changes the direction of light at a boundary.
The incident and reflected rays remain in the same medium.
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Angles are measured from the normal to the surface.
The incident ray, reflected ray, and normal lie in one plane.
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Refraction occurs because wave speed changes across a boundary.
The frequency remains constant while wavelength changes.