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  4. A mass of 4 kg suspended from a spring of force constant 800 N m-1 executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in ms-2) of the mass is

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Q. A mass of $4\, kg$ suspended from a spring of force constant $800 \,N\, m^{-1}$ executes simple harmonic oscillations. If the total energy of the oscillator is $4\, J$, the maximum acceleration (in $ms{-2}$) of the mass is

JIPMERJIPMER 2013Oscillations

Solution:

Here ,$m = 4 \, kg , k = 800 \, N \, m^{-1} , E = 4 \, J$
In SHM, total energy is $E = \frac{1}{2} kA^2$
$\therefore \:\:\:\: 4 = \frac{1}{2} \times 800 \times A^2$
$ A = \frac{1}{10} m = 0.1 m$
Maximum acceleration, $a_{max} = \omega^2 A$
$= \frac{k}{m}A = \frac{800 N m^{-1}}{4 kg} \times0.1 m \left( \because \omega = \sqrt{\frac{k}{m}}\right)$
$ = 20 \, m \, s^{-1}$