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

Book Online Demo

Check Your IQ

Try Test

Courses

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

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 60.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

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

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

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

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

Watch 3-min video & get full concept clarity