An object of mass 8.0 kg is attached to an ideal massless spring and llowed to hang in the Earth's gravitational field. The spring stretches 3.6 em before it reaches its equilibrium position. If this system is allowed to oscillate, what will be its frequency? O 0.0045 Hz O 2.1 Hz O 0.67 Hz O 2.6 Hz
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- How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 79.5 N/m so that it will oscillate at 1.00 Hz? O 39.5 kg O 2.00 kg O 1.00 kg O 1.56 kgThe pendulum as shown makes small oscillations at cycle frequency f = 0.580 Hz . Next, the cord is replaced with a new cord one-third the length of the original and the mass of the pendulum is tripled. What is the frequency of small oscillations of the pendulum now? 0.580 Hz 0.193 Hz 1.74 Hz 1.00 Hz 0.335 HzThe position of a particle is given by the expression x = 2.00 cos (2.00nt + π/4), where x is in meters and t is in seconds. (a) Determine the frequency. Hz (b) Determine period of the motion. S (c) Determine the amplitude of the motion. m (d) Determine the phase constant. rad (e) Determine the position of the particle at t = 0.350 s. m
- While fishing for catfish, a fisherman suddenly notices that thebobber (a floating device) attached to his line is bobbing up anddown with a frequency of 1.8 Hz. What is the period of the bobber’s motion?Find the frequency value of this question.To drive a spring with k = 120 and a mass of 3 kg (attached) in resonance at what frequency should the energy be applied? 4 Hz 1 Hz 2 Hz 2 Hz
- A mass of 1.64 Kg is connected to a spring of spring constant 9.02 N/m. An oscillation is started by pulling the mass to the right to amplitude 0.779 m before release and the oscillator moves in air. The oscillation decays to 16.7% of the original amplitude in 63.5 seconds. a. What would the position of the oscillation be 29.63 seconds after release?A 0.500 kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the path at a constant frequency of 2.20 rev/sec. At the moment the stone is overhead, the stone is released. The velocity of the stone when it leaves the circular path is Multiple Choice 10.4 m/s horizontal. 10.4 m/s vertical. 22.0 m/s horizontal. 22.0 m/s vertical. 31.4 m/s horizontal.A mass of 0.407 kilograms is oscillating horizontally on a spring. Its position is given by the equation x = [0.108 sin (9.24t + 0.82) + 0.683] m. a. What is the spring constant of the spring? Include units in your answer. b. The velocity of the mass is another sinusoid, which also can be modeled using the equation v = [A sin (Bt + C) + D] m/s. What are those values? Do not bother with units.A: B: C: (between 0 and 2?)D: c. The acceleration of the mass is yet another sinusoid, which again can be modeled using the equation a = [A sin (Bt + C) + D] m/s/s. What are those values of A and C? Do not bother with units. Hint: For A, consider Newton's second law of motion. A: C: (between 0 and 2?)Note that the values of B and D for the acceleration equation are the same as the values of B and D for the velocity equation. PLEASE ANSWER IN HANDWRITING THANKS!!!!
- The position of a particle is given by the expression x = 6.00 cos (4.00nt + 2π/5), where x is in meters and t is in seconds. (a) Determine the frequency. Hz (b) Determine period of the motion. S (c) Determine the amplitude of the motion. m (d) Determine the phase constant. rad (e) Determine the position of the particle at t = 0.350 s. mthe position of a mass oscillating on a spring is given by x=(7.8cm)cos(2pi/0.68s). a) what is the frequency of this motion? b)When is the mass first at the position x=-7.8cm?The position of a particle is given by the expression x = 6.00 cos (6.00xt + T/2), where x is in meters and t is in seconds. (a) Determine the frequency. Hz (b) Determine period of the motion. (c) Determine the amplitude of the motion. (d) Determine the phase constant. rad (e) Determine the position of the particle at t = 0.270 s.