touchNOF“ 72| 1:09 PMPHYS220_Spring-20-21...PHYS220 _ Spring/20-21 _ Practice Sheet-1 _ Keys1) A0.1 kg object oscillates as a simple harmonic motion along thex -axis with a frequency f = 3.185 Hz.At a position x1, the object has a kinetic energy of 0.7 J and a potential energy 0.3 J. The amplitude ofoscillation, A, is:(a) 0.12 m(b)0.22 m(c) 0.31 m(d)0.42 m2) A block of mass m is attached to a spring with force constant, k and oscillate at a frequency f. If themass is changed to m' = m/2, and the spring force is changed to k' = 2k, then the new frequency f' ofthe oscillation would be,(a) f' = 2f(b)f' = f(c) f' = f/2(d)/' = 4fThe following given is for questions 3 and 4:A block of mass m = 2 kg is attached to a spring with spring constant k = 200 N/m, and set tooscillates on a frictionless horizontal surface. At time t = 0 its position is x, = 0 and its velocity isvo = +5 m/s.3) Which of the following is true about the oscillation amplitude and the phase constant, p:(a) A = 0.5, o = a/2(b)A = 0.5, 0 = -7/2(c) A = 0.25, 9 =x/2(d)A = 0.25, 9 = -x/24) When the block's velocity has the value v = v,may/2, then its position has the possible values of:(a) x = +(3/4)A(b)x = ±A/2(c) x = ±(v3/2)A(d)x = ±(vZ/2)A5) A block-spring system oscillates in a simple harmonic motion on a frictionless horizontal table. Itsdisplacement varies with time according to x(t) = 0.2 cos(2t – n/4). The earliest time the particlereaches position x = 0.1 m is(а) t -24(b) t =S2447m(c) t =247m(d) t =126) Consider a bob attached to a vertical string. At t = 0, the bob is displaced to a new position where thestring makes -8° with the vertical and is then released from rest. If the angular displacement e(t) iswritten as a cosine function, then the phase constant o is:(а) -п/2(b)0(c) +n/2(d)n7) A bob attached to a cord is moved to the right where its vertical position is 1.05 cm above thesauilibrium position and is then given an initial speed of 0the values of the mavimum

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Description: touchNOF“ 72| 1:09 PMPHYS220_Spring-20-21...PHYS220 _ Spring/20-21 _ Practice Sheet-1 _ Keys1) A0.1 kg object oscillates as a simple harmonic motion along thex -axis with a frequency f = 3.185 Hz.At a position x1, the object has a kinetic energy of

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1) A0.1 kg object oscillates as a simple harmonic motion along the x-axis with a frequency f = 3.185 Hz. At a position x, the object has a kinetic energy of 0.7 J and a potential energy 0.3 J. The amplitude of oscillation, A, is: (a) 0.12 m (b)0.22 m (c) 0.31 m (d)0.42 m

1) A0.1 kg object oscillates as a simple harmonic motion along the x-axis with a frequency f = 3.185 Hz. At a position x, the object has a kinetic energy of 0.7 J and a potential energy 0.3 J. The amplitude of oscillation, A, is: (a) 0.12 m (b)0.22 m (c) 0.31 m (d)0.42 m

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Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A 0.1 object oscillates as a simple harmonic motion along the x-axis with a frequency f=3.185 hz. At a position x the object has a kinetic energy 0.7 j and a potential energie 0.3 j . What is the amplitude of oscillationA?
A 0.1 kg object oscillates as a simple harmonic motion along the xx −axis with a frequency ff = 3.185 Hz.At a position xx1, the object has a kinetic energy of 0.7 J and a potential energy 0.3 J. The amplitude ofoscillation, AA, is:
To double the total energy of a mass oscillating at the end of a spring, we need to Select one: O a. increase the amplitude by 4 and decrease the angular frequency by√2 O b. increase the amplitude by √2 O c. increase the angular frequency by 2. O d. increase the angular frequency by √2 e. increase the amplitude by 2
A O1 26% 0 3:38 pm live6.pdf ビy 1) A 0.1 kg object oscillates as a simple harmonic motion along the x -axis with a frequency f = 3.185 Hz. At a position x,, the object has a kinetic energy of 0.7 J and a potential energy 0.3 J. The amplitude of oscillation, A, is: f- 3.18 sH 3 (a) 0.12 m m=0.l Ka (b) 0.22 m KE= 0.78 PE =0.3J. (c) 0.31 m (d) 0.42 m A -7: E= XE +PE = o.7+ 0.3 =1J= KA". => A?= 2-2 K m=0.22 K = mw? = (0.1)(20) * : 40. => w : 20 radls 2) A block of mass m is attached to a spring with force constant, k and oscillate at a frequency f. If the mass is changed to m' = m/2, and the spring force is changed to k' = 2k, then the new frequency f' of the oscillation would be, (a)f = 2f / m'- m K': 2K (b)f = f (c)f' = f/2 (d)f" = 4f wiVE = = 2 w. in = The following given is for questions 3 and 4: A block of mass m = 2 kg is attached to a spring with spring constant k = 200 N/m, and set to oscillates on a frictionless horizontal surface. At time t = 0 its position is xo = 0 and its…
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3) OK, once again we have a pendulum, this time of length 1.94 m, which you release from rest at an angle of 59.1 degrees to the vertical. What will be the speed of the pendulum at the instant it reaches an angle of 29.6 degrees above the vertical? Question 2 options: 6.88 m/s 3.68 m/s 4.30 m/s 2.23 m/s 4) This, the length of the pendulum is 2.17 m. Now you start with the pendulum at 10.2 degrees with respect to the vertical, but rather than releasing it from rest, you give it a push downward. It swings to the other side, and reaches a maximum angle of 34.0 degrees with respect to the vertical. What must have been the initial speed of the pendulum just after you pushed it? 2.70 m/s 0.82 m/s 3.51 m/s 2.57 m/s
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A 0.1 kg object oscillates as simple harmonic motion along the x axis with a frequency f=3.185 Hz . At a position x1 the object has a kinetic energy of 0.7J and a potential energy of 0.3J . Calculate the amplitude of ascillation A
1) A 0.1 kg object oscillates as a simple harmonic motion along the x -axis with a frequency f = 3.185 Hz. At a position x1, the object has a kinetic energy of 0.7 J and a potential energy 0.3 J. The amplitude of oscillation, A, is: (a) 0.12 m (b)0.22 m (c) 0.31 m (d)0.42 m 2) A block of mass m is attached to a spring with force constant, k and oscillate at a frequency f. If the mass is changed to m' = m/2, and the spring force is changed to k' = 2k, then the new frequency f' of the oscillation would be, (a) f' = 2f (b)f' = f (c) f' = f/2 (d)f' = 4f %3D
1) A0.1 kg object oscillates as a simple harmonic motion along the x-axis with a frequency f = 3.185 Hz. At a position x,, the object has a kinetic energy of 0.7 J and a potential energy 0.3 J. The amplitude of oscillation, A, is: (a) 0.12 m (b)0.22 m (c) 0.31 m (d) 0.42 m