The proper period of an oscillator is 1.50000 s. In an environment with damping proportional to the speed the period increases to 1,50015 s. Find the number of cycles N so that the energy is reduced by 25% of its initial value and the quality factor of the system.
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The proper period of an oscillator is 1.50000 s. In an environment with damping proportional to the speed the period increases to 1,50015 s. Find the number of cycles N so that the energy is reduced by 25% of its initial value and the quality factor of the system.
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- H In a damped oscillator, let m = 250 g, k=85 N/m, and b=0.070 kg/s. In how many periods of oscillation the mechanical energy of the oscillator drop to one-half of its initial value?A 190 g mass suspended from a spring oscillates in simple harmonic motion. The mass completes 2 cycles every second. Calculate the spring constant.Calculate the velocity of a simple harmonic oscillator with amplitude of 13.5 cm and frequency of 5 Hz at a point located 5 cm away from the equilibrium position. Give your answer in Sl units.
- A 0.250-kg glider on a horizontal air track is attached to an ideal spring. When the glider is pulled 0.400 m from its equilibrium position and released from rest, it oscillates with an angular frequency of 11.0 rad/s. (Ignore friction.) (a) Find the force constant of the spring. (b) Find the period of the oscillations. (c) Find the maximum speed of the mass.Because of dissipative forces, the amplitude of an oscillator decreases 3.96% in 10 cycles. By what percentage does its energy decrease in ten cycles?a particle of mass m makes simple harmonic motion with force constant k around the equilibrium position x = 0. Calculate the valu of the amplitude and the phase constant if the initial conditions are x(0) = 0 and v(0) = -Vi (vi > 0). Draw x-t, v - t and a - t figures for one complete cycle of oscillations.
- Consider a critically damped oscillator with w=y and mass m that is driven by force Fa cos(wat). ▼ Part A - What is the amplitude of the steady-state oscillation at wa=0? Answer symbolically. You can type y as \gamma. You can type Fa as F_d Ao = Submit IVE ΑΣΦ Fd k Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Wd = Part B - The maximum amplitude of oscillation, Ao, occurs for wa = 0. At what driving frequency does the amplitude equal to Ao/2? Answer symbolically. You can type y as \gamma. You can type Fa as F_d Submit ? IVE ΑΣΦ Request Answer ?Because of dissipative forces, the amplitude of an oscillator decreases 4.94% in 10 cycles. By what percentage does its energy decrease in ten cycles?A block of mass (3 kg) lays on a frictionless plane and is attached to a horizontal spring with a spring constant (8 N/m). The spring is attached to a nearby vertical wall. The minimum distance of the block from the wall during its oscillations is 0.8 m. The maximum distance of the block from the wall during its oscillations is 1.2 m. If we start a stopwatch (t = 0) when the block is at its closest point to the vertical wall, how far the block from the wall after a time t = 5/4 T? A=0.2 m V= 0.326 T= 0.383s
- A 236 g ball is tied to a string. It is pulled to an O angle of 7.8 and released to swing as a pendulum. A student with a stopwatch finds that 13 oscillations take 18 s. You may want to review (Pages 405 - 407). How long is the string? Express your answer to two significant figures and include the appropriate units. L = μÅ Value Ć Units ?A mass of 16 kilograms is suspended from a spring with spring constant 7 Kg/sec^2. The damping constant is 2 Kg/sec and the spring is subject to an external force F(t)= 3cos(9t) Newtons. The motion of the mass eventually settles down to a periodic oscillation. What is the period (in sec) of this periodic oscillation? 2 9.A 5 kg mass is hung on a spring with a stiffness of 70 N/m. Find the oscillation response of this mass as a function of time, if x0= 1 m and v0= - 0.45 m/s. x0 is the initial displacement and vo the initial speed. x = cos (3.74 t) - 0.12 sin(3.74 t) x = cos (14 t) - 0.12 sin(14 t) x = - 0.12 cos (3.74 t) + sin(3.74 t) x = - 0.12 sin(3.74 t) + cos (14 t)