An object of mass 0.2 kg. executes simple harmonic motion along the x-axis with a frequency of 25π Hz at the position x = 0.04 m, the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillator is _______ cm.

TE of oscillator TE=12mω2A2ω=2πT=4πT2=4π2υ2=ω2TE=12m4π2υ22A2=m=0.2kg, υ=25πTE=PE+KE=0.5+0.4=0.9TE=12m4π2υ2A2⇒A2=2TE4π2mυ2A=6cm

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An object of mass 0.2 kg. executes simple harmonic motion along the x-axis with a frequency of $(\frac{25}{π})$ Hz at the position x = 0.04 m, the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillator is _______ cm.

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Detailed Solution

TE of oscillator $T E = \frac{1}{2} m ω^{2} A^{2}$

$ω = \frac{2 π}{T} = \frac{4 π}{T^{2}} = 4 π^{2} υ^{2} = ω^{2}$

$T E = \frac{1}{2} m 4 π^{2} υ^{2} 2 A^{2}$

$= m = 0.2 k g, υ = \frac{25}{π}$

$T E = P E + K E = 0.5 + 0.4 = 0.9$

$T E = \frac{1}{2} m 4 π^{2} υ^{2} A^{2} \Rightarrow A^{2} = \frac{2 T E}{4 π^{2} m υ^{2}}$

$A = 6 c m$

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