A particle with in SHM has an equation
X=0.75Sin65t cm. What is:
A. the frequency of the oscillation
B. The amplitude of the oscillation
C. The time period of the oscillation
x=0.75sin65tx=0.75\sin65tx=0.75sin65t
x=Asinωtx = A\sin\omega tx=Asinωt
A=75cm (amplitude)A = 75 cm\text{ (amplitude)}A=75cm (amplitude)
ω=65 (cyclic frequency)\omega = 65\text{ (cyclic frequency)}ω=65 (cyclic frequency)
A.f=ω2π=652π=10.35HzA.f = \frac{\omega}{2\pi}= \frac{65}{2\pi}= 10.35 HzA.f=2πω=2π65=10.35Hz
B.A=75cm=0.75mB. A = 75 cm = 0.75 mB.A=75cm=0.75m
C.T=1f=110.35=0.097sC.T = \frac{1}{f}=\frac{1}{10.35}=0.097 sC.T=f1=10.351=0.097s
Answer:\text{Answer:}Answer:
A.f=10.35HzA.f = 10.35 HzA.f=10.35Hz
B.A=0.75mB. A = 0.75 mB.A=0.75m
C.T=0.097sC.T = 0.097 sC.T=0.097s
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