qz10sol_3705s11 - position function: v(t) = dx/dt = - x m...

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TA: Tomoyuki Nakayama Monday, April 11, 2011 PHY 2048: Physic 1, Discussion Section 3705 Quiz 10 (Homework Set #12) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ 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 = 40.0 cm/s. a) What is the angular frequency for the harmonic oscillator? The velocity of the oscillator is the time derivative of the
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Unformatted text preview: position function: v(t) = dx/dt = - x m ω sin( ω t + φ ) It takes the maximum value of x m ω . Therefore, the maximum velocity is given by v m = x m ω . According to the graph, the maximum velocity is 25 cm/s. The angular frequency is ω = v m /x m = 10 rad/s b) What is the phase constant (from 0 to 2 π rad) for the harmonic oscillator? The velocity and acceleration of the oscillator are v(t) = dx/dt = - v m sin( ω t + φ ) , a(t) = dv/dt = -a m cos( ω t + φ ) At time t = 0, the velocity takes the value v s = 40 cm/s. The velocity function yields v s = -v m sin( φ ) ⇒ φ = 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( φ ) > 0 π /2< φ <3 π /2 Therefore, the phase constant is 4.07 rad....
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.

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