// Numbas version: exam_results_page_options {"name": "Shifted Simple Harmonic Motion Graph", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"extensions": ["geogebra"], "tags": [], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

This uses an embedded Geogebra SHM graph with coefficients set by NUMBAS.

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The depth of water, $h$ metres, of a harbour at time $t$ hours, can be modelled as simple harmonic motion about a fixed depth.

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{geogebra_applet('https://ggbm.at/njwfpvb7', defs, [])}

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SHM origin, $h = $ [[0]] metres

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Maximum depth, $h_{max} = $ [[1]] metres

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Minimum depth, $h_{min} = $ [[2]] metres

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Which of the following best describes the depth of water after $t$ hours?

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There are two high tides in a 25 hour period.  Calculate the angular frequency for the harbour depth:

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Angular frequency, $\\omega = $ [[0]] rad/hour

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Sets the internal values of y = a*cos(wx+c)+d

", "name": "defs"}, "b": {"definition": "random(1..10)", "group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "b"}}, "advice": "

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(c) Two high-tides in 25 hours means the period is half of that value:  1 high tide per 12.5 hours. Therefore \\[ P = 12.5 \\text{ hours} \\] Angular frequency, $\\omega$, is defined as \\[ \\omega = \\frac{2\\pi}{12.5} =\\frac{4\\pi}{25} \\, \\mathrm{rad/hr} \\] where $P$ is the period of oscillation.

", "type": "question", "contributors": [{"name": "Adrian Jannetta", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/164/"}]}]}], "contributors": [{"name": "Adrian Jannetta", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/164/"}]}