An object of mass 0.2 kg executes S.H.M along the x-axis with a frequency of 25πHz at the point x = 0.04m the object has K.E. of 0.5J and P.E of 0.4J. The amplitude of oscillation in ‘mm’ is

T.E=12mω2A2=m⋅2π2n2A2T⋅E=P⋅E+K⋅E=0.9J0.9=0.2×2×π2×25×25π2×A2∴A2=0.90.2×2×25×25=94×25×25A=350=0.06 m ∴A=60 mm

An object of mass 0.2 kg executes S.H.M along the x-axis with a frequency of $\frac{25}{π} Hz$ at the point x = 0.04m the object has K.E. of 0.5J and P.E of 0.4J. The amplitude of oscillation in ‘mm’ is

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$T . E = \frac{1}{2} {mω}^{2} A^{2} = m \cdot 2 π^{2} n^{2} A^{2}$

$T \cdot E = P \cdot E + K \cdot E = 0 .9 J$

$0 .9 = 0 .2 \times 2 \times π^{2} \times \frac{25 \times 25}{π^{2}} \times A^{2}$

$∴ A^{2} = \frac{0 .9}{0 .2 \times 2 \times 25 \times 25} = \frac{9}{4 \times 25 \times 25}$

$A = \frac{3}{50} = 0 .06 m ∴ A = 60 mm$

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