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qz10sol_6942s11

# qz10sol_6942s11 - spring force by the right spring...

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TA: Tomoyuki Nakayama Monday, April 11, 2011 PHY 2048: Physic 1, Discussion Section 6942 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. ________________________________________________________________________________ In the figure below right, two light springs are joined and connected to a block of mass 40.0 kg that is set oscillating over a frictionless floor. The springs have spring constants k 1 = 200 N/m and k 2 = 500 N/m respectively. a) What is the ratio of the elongation (or compression) of spring 1 to that of spring 2? Since the springs are light, we can neglect the masses of the springs. Two forces are exerted on the left spring: A force by the block and
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Unformatted text preview: spring force by the right spring. Newton’s 2 nd law applied to the left spring yields 0 = k 1 Δ x 1 – k 2 Δ x 2 ⇒ Δ x 1 / Δ x 2 = k 2 /k 1 = 2.5 Note the force by the block on the spring has the same magnitude as the force on the block by the spring. b) What is the frequency (in Hz) of the oscillations? The block is attached to the left spring. Therefore, only the left spring exert spring force on the block. Applying Newton’s 2 nd law to the block, we get ma = -k 1 Δ x 1 We express Δ x 1 in terms of the displacement of the block. x = Δ x 1 + Δ x 2 = Δ x 1 + Δ x 1 /2.5 Δ x 1 = (2.5/3.5)x Therefore, the acceleration of the block is proportional to the displacement of the block: a = (2.5 k 1 /3.5m)x The proportionality constant is equal to the square of the angular frequency. The frequency is f = ω /(2 π ) = √ (2.5 k 1 /3.5m)//(2 π ) = 0.301 Hz...
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