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 = 0.4 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 /0.4 Δ x 1 = (0.4/1.4)x Therefore, the acceleration of the block is proportional to the displacement of the block: a = (0.4k 1 /1.4m)x The proportionality constant is equal to the square of the angular frequency. The frequency is f = ω /(2 π ) = √ (0.4 k 1 /1.4m)//(2 π ) = 0.301 Hz...
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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.
- Spring '08