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# hw12 - Homework 12 Solutions(partial Problem 14 26 The mass...

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Unformatted text preview: Homework 12 Solutions (partial) Problem 14 . 26 The mass is struck with an impulse J . This means that: Δ p = J ⇒ mv- 0 = J ⇒ v = J m Where v is the velocity of the mass right after it was struck (we can do this because at that moment there is no external force). Next, we know the general solution to for a simple harmonic oscillator is: x ( t ) = a cos( ωt ) + b sin( ωt ) ⇒ v ( t ) =- aω sin( ωt ) + bω cos( ωt ) Now, since x (0) = 0, this implies that a = 0 in the above equation. Similarly, since v (0) = v = J m , we see that: bω = J m = ⇒ b = J ωm Where ω = radicalBig k m is the angular frequency of oscillation. Hence, we find that: x ( t ) = J m radicalBig k m sin parenleftBigg radicalbigg k m t parenrightBigg = J √ km sin parenleftBigg radicalbigg k m t parenrightBigg Problem 14 . 27 First of all, write down x ( t ) and v ( t ): x ( t ) = A cos( ωt ) v ( t ) =- Aω sin( ωt ) a) The amplitude is just the coefficient of cos( ... ): A 1 b) The period is given in terms of ω as: T = 2 π ω . Since the frequency is 1 /T , this implies that: f = 1 T = ω 2 π c) The easiest way to find the total energy is to find the maximum kinetic energy. Since at the point of maximum kinetic energy, the potential energy is zero, and I’ve found the total energy. First note that since sin( ... ) ≤ 1 that: v ≤ Aω . Hence: E tot = 1 2 mv 2 max = 1 2 m ( ωA ) 2 d) Here, I have to find out what the velocity is in terms of the position, x . Then I can rewrite the kinetic energy in terms of x as well. Remember that cos 2 θ + sin 2 θ = 1 so that: v 2 = (- Aω ) 2 sin 2 ( ωt ) = A 2 ω 2 (1- cos 2 ( ωt )) = ω 2 [ A 2- ( A 2 cos 2 ( ωt ) ) ] = ω 2 [ A 2- x 2 ] Now I can get the kinetic energy at the point x : KE = 1 2 mv 2 = 1 2 mω 2 ( A 2- x 2 ) e) Finally, to get the potential energy, I remember that E tot = KE + U , so that: U = E tot- KE . Hence: U = 1 2 mω...
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hw12 - Homework 12 Solutions(partial Problem 14 26 The mass...

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