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A spring of force constant 1200 Nm^(-1) ...

A spring of force constant `1200 Nm^(-1)` is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of spring, pulled sideways to a distance of 2.0 cm and released.
(a) What is the frequency of oscillation of mass ?
(b) What is its maximum acceleration ?
(c) What is its maximum speed ?

Text Solution

Verified by Experts

`k = 1200 Nm^(-1), m = 3.0 kg`
` a = 2 xx10^(-2)m`
(a) Frequency of oscillation `v=1/(2pi)sqrt(k/m)=1/(2xx3.14) sqrt((1200)/3)`
`v = 3.18 ` Hertz
(b) Maximum acceleration `alpha_(max)=omega^2a=ak/m=2xx10^(-2)xx1200/(3.0) = 8 ms^(-2)`
(c) Maximum velocity `alpha_(max)=aomega=2xx10^(-3)xxsqrt((1200)/3)=0.4 ms^(-1)`
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