Question

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

• A. 5
• B. 15
• C. 45
• D. 20

Answer 1

Answer: (D)

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