A simple pendulum has length 1 m. Using g=9.8 m/s², what is its period approximately?
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Use T=2π√(L/g).
T=2π√(1/9.8)≈2.01 s.
Practice MDCAT Physics questions with answers and explanations.
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Use T=2π√(L/g).
T=2π√(1/9.8)≈2.01 s.
Choose an option to check your answer.
For a spring oscillator, ω=√(k/m).
√(200/0.5)=√400=20 rad/s.
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Period is T=2π/ω.
T=2π/10≈0.628 s.
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Speed is distance divided by time.
8/0.4 = 20 m/s.
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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.
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In-phase displacements add constructively.
The resultant amplitude is 3+2 = 5 cm.
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Opposite-phase amplitudes subtract.
The resultant amplitude is |5-3| = 2 cm.
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Greater wave amplitude generally means greater intensity.
Perceived loudness is also frequency-dependent.
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For the fundamental, the string length is half a wavelength.
λ=2L=2.4 m.
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Approach raises observed frequency, while separation lowers it.
The effect occurs for sound and electromagnetic waves.
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The fundamental wavelength is 2L=1.5 m.
F=v/λ=300/1.5 = 200 Hz.
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The period is measured in seconds.
It is the reciprocal of frequency.