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A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released. Determing. (a) the frequency of oscillation of the mass. (b) the maximum acceleration of the mass. (c) the maximum speed of the mass. |
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Answer» Here, `k=1200Nm^(-1),m=3.0kg, a=2.0cm=0.2m` (a) Frequency, `v=(1)/(T)=(1)/(2pi)sqrt((k)/(m))=(1)/(2xx3.14)sqrt((1200)/(3))=3.2s^(-1)` (b) Acceleration, `A=omega^(2)y=(k)/(m)y` Acceleration will be maximum when y is maximum, i.e., `y=a` `:. ` Max. acceleration, `A_(max)=(ka)/(m)=(1200xx0.02)/(3)=8ms^(-2)` (c) Max. speed ofo the mass will be when it is passing through the mean position, which is given by `V_(max)=aomega=asqrt((k)/(m))=0.02xxsqrt((1200)/(3))=0.4ms^(-1)` |
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