An object of mass 0-2 kg is in SHM along x-axis with a frequency of 25- hertz- At the position x -0-04m- it has KE of 0-5 J and PE of 0-4 J- its amplitude is-

The correct option is A 0.06 m TE=PE+KE TE=12m ω^2A^2 ω=2π t =2π×25π =50 ∴12× 0.2× (50)^2× A^2=0.5+0.4 A^2=9(50)^2 ...

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Standard XII

Physics

COE in SHM

An object of ...

Question

An object of mass $0.2$ kg is in SHM along x-axis with a frequency of $\frac{25}{π}$ hertz. At the position $x = 0.04$ m, it has KE of $0.5$ J and PE of $0.4$ J, its amplitude is:

0.06

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Solution

The correct option is A $0.06$ m
$T E = P E + K E$
$T E = \frac{1}{2} m ω^{2} A^{2}$
$ω = 2 π t$
$= 2 π \times \frac{25}{π}$
$= 50$
$∴ \frac{1}{2} \times 0.2 \times (50)^{2} \times A^{2} = 0.5 + 0.4$
$A^{2} = \frac{9}{(50)^{2}}$
$A = \frac{3}{50}$
$A = 0.06 m$

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An object of mass 0.2 kg is in SHM along x-axis with a frequency of 25π hertz. At the position x=0.04m, it has KE of 0.5 J and PE of 0.4 J, its amplitude is:

The correct option is A 0.06 mTE=PE+KETE=12mω2A2ω=2πt=2π×25π=50∴12×0.2×(50)2×A2=0.5+0.4A2=9(50)2A=350A=0.06m

An object of mass $0.2$ kg is in SHM along x-axis with a frequency of $\frac{25}{π}$ hertz. At the position $x = 0.04$ m, it has KE of $0.5$ J and PE of $0.4$ J, its amplitude is:

The correct option is A $0.06$ m
$T E = P E + K E$
$T E = \frac{1}{2} m ω^{2} A^{2}$
$ω = 2 π t$
$= 2 π \times \frac{25}{π}$
$= 50$
$∴ \frac{1}{2} \times 0.2 \times (50)^{2} \times A^{2} = 0.5 + 0.4$
$A^{2} = \frac{9}{(50)^{2}}$
$A = \frac{3}{50}$
$A = 0.06 m$