Question #72634

The amplitude of an oscillator is 8 cm and it completes 100 oscillations in 80s.
i) Calculate its time period and angular frequency. ii) If the initial phase is π/4, write
expressions for its displacement and velocity. iii) Calculate the values of maximum
velocity and acceleration.
ii0 body of mass 0.15 kg executes SHM described by the equation
x(t) = 2sin(πt + π )4/
where x is in meters and t is in seconds. i) Determine the amplitude and time period
of the oscillation. ii) Calculate the initial values of displacement and velocity.
iii) Calculate the values of time when the energy of the oscillator is purely kinetic

Expert's answer

Answer on Question #72634-Physics-Other

The amplitude of an oscillator is 8cm8\mathrm{cm} and it completes 100 oscillations in 80s.

i) Calculate its time period and angular frequency.

ii) If the initial phase is π/4\pi /4, write expressions for its displacement and velocity.

iii) Calculate the values of maximum velocity and acceleration.

Solution

i)


T=80100=0.8s.T = \frac{80}{100} = 0.8\,s.ω=2πT=2π0.8=7.85rads.\omega = \frac{2\pi}{T} = \frac{2\pi}{0.8} = 7.85\,\frac{rad}{s}.


ii)


x(m)=0.08sin(5π2t+π4)x(m) = 0.08 \sin \left(\frac{5\pi}{2}t + \frac{\pi}{4}\right)v(ms)=0.08(2π0.8)cos(5π2t+π4)=π5cos(5π2t+π4)v\left(\frac{m}{s}\right) = 0.08 \left(\frac{2\pi}{0.8}\right)\cos \left(\frac{5\pi}{2}t + \frac{\pi}{4}\right) = \frac{\pi}{5}\cos \left(\frac{5\pi}{2}t + \frac{\pi}{4}\right)


iii)


V=π50.63ms.V = \frac{\pi}{5} \approx 0.63\,\frac{m}{s}.a=π5(2π0.8)=4.9ms2.a = \frac{\pi}{5} \left(\frac{2\pi}{0.8}\right) = 4.9\,\frac{m}{s^2}.


Body of mass 0.15kg0.15\,\mathrm{kg} executes SHM described by the equation


x(t)=2sin(πt+π/4)x(t) = 2\sin(\pi t + \pi/4)


where xx is in meters and tt is in seconds.

i) Determine the amplitude and time period of the oscillation.

ii) Calculate the initial values of displacement and velocity.

iii) Calculate the values of time when the energy of the oscillator is purely kinetic

Solution

i)


A=2m.A = 2\,m.T=2ππ=2 s.T = \frac {2 \pi}{\pi} = 2 \text{ s}.


ii)


x(0)=2sin(π4)=2(22)=21.41 m.x(0) = 2 \sin \left(\frac {\pi}{4}\right) = 2 \left(\frac {\sqrt {2}}{2}\right) = \sqrt {2} \approx 1.41 \text{ m}.v(0)=2(π)cos(π4)=2(π)(22)=π24.44ms.v(0) = 2(\pi) \cos \left(\frac {\pi}{4}\right) = 2(\pi) \left(\frac {\sqrt {2}}{2}\right) = \pi \sqrt {2} \approx 4.44 \frac {m}{s}.


iii)


x(t)=2sin(πt+π4)=0x(t) = 2 \sin \left(\pi t + \frac {\pi}{4}\right) = 0πt+π4=πn\pi t + \frac {\pi}{4} = \pi nt=n14,n=1,2,3,t = n - \frac {1}{4}, n = 1, 2, 3, \dotst=34 s,134 s,234 st = \frac {3}{4} \text{ s}, 1 \frac {3}{4} \text{ s}, 2 \frac {3}{4} \text{ s} \dotst=0.75 s,1.75 s,2.75 st = 0.75 \text{ s}, 1.75 \text{ s}, 2.75 \text{ s} \dots


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