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A spring having a spring constant 1200Mn^(-1) is mounted on a horizontal table. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 0.02 m and released. Determine (i) frequency of oscillations (ii) maximum acceleration of the mass and (iii) the maximum speed of the mass. |
Answer» Solution :Given, `K=1200Nm^(-1), m=3kg, y=0.02 m` (i) Frequency of oscillation, `F=(1)/(2pi)sqrt((k)/(m))` i.e. `f=((1)/(2xx3.142))sqrt((1200)/(3))=(10)/(3.142)=((10)/(PI))Hz` (ii) maximum acceleration of the mass, `a_("max")=-omega^(2)y` where `omega=(2pi)/(T)=2pif` `"i.e."omega=2pixx(10)/(pi)` hence `a_("max")=-20xx20xx0.02` `a_("max")=8ms^(-2)` (III) maximum speed of the mass, `v_("max")=omegaA=2xx0.02 =0.4ms^(-1)`. |
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