qz11sol_0132Hf10

# qz11sol_0132Hf10 - ) It takes the maximum value of x m ....

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TA: Tomoyuki Nakayama Monday, November 22, 2010 PHY 2048: Physic 1, Discussion Section 0132H Quiz 11 (Homework Set #13) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The figure below right shows the velocity function v(t) for a simple harmonic oscillator. The oscillator’s position function x(t) has the form x = x m cos( ω t + φ ), where x m = 5.00 cm The vertical axis scale of the figure is set by v s = 20.0 cm/s. a) What is the angular frequency for the harmonic oscillator? The velocity function of the oscillator is the time derivative of the position function. v(t) = dx/dt = - x m ω sin( ω t +
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Unformatted text preview: ) It takes the maximum value of x m . Therefore, the maximum velocity is given by v m = x m . According to the graph, the maximum velocity is 25 cm/s. The angular frequency is = v m /x m = 5 rad/s b) What is the phase constant (from 0 to 2 rad) for the harmonic oscillator? The velocity and acceleration of the oscillator are v(t) = dx/dt = - v m sin( t + ) , a(t) = dv/dt = -a m cos( t + ) At time t = 0, the velocity takes the value v s = 20 cm/s. The velocity function yields v s = -v m sin( ) = sin-1 (-v s /v m ) = 4.07 rad, 5.36rad At time t = 0, the acceleration (i.e. the slope of the graph) is positive. The acceleration function yields -a m cos( ) &amp;gt; 0 /2&amp;lt; &amp;lt;3 /2 Therefore, the phase constant is 4.07 rad....
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## This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.

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