<p><strong>The potential energy of a particle of mass 4 kg in motion along the x-axis is given by <math><mrow><mi>U</mi><mo>=</mo><mn>4</mn><mrow><mo>(</mo><mrow><mn>1</mn><mo>−</mo><mi>c</mi><mi>o</mi><mi>s</mi><mn>4</mn><mi>x</mi></mrow><mo>)</mo></mrow><mi>J</mi><mtext> </mtext><mo>.</mo></mrow></math> The time period of the particle for small oscillation (sin = ) is <math><mrow><mrow><mo>(</mo><mrow><mfrac><mrow><mi>π</mi></mrow><mrow><mi>K</mi></mrow></mfrac></mrow><mo>)</mo></mrow><mi>s</mi><mo>.</mo></mrow></math> The value of K is_______.</strong></p>

m = 4kgU = 4 (1 – cos 4x) JF=−∂υ∂x=−4×4sin4x∴a=−164sin4x=−4sin4x=−16x⇒a=−16xω2=16⇒ω=4=2πT⇒T=2π4=π2seck = 2

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Physics Class 12th Oscillations Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

The potential energy of a particle of mass 4 kg in motion along the x-axis is given by $U = 4 (1 - c o s 4 x) J .$ The time period of the particle for small oscillation (sin = ) is $(\frac{π}{K}) s .$ The value of K is_______.

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Vishal Baghel | Contributor-Level 10

3 months ago

m = 4kg

U = 4 (1 – cos 4x) J

$F = - \frac{\partial υ}{\partial x} = - 4 \times 4 s i n 4 x$

$∴ a = - \frac{16}{4} s i n 4 x = - 4 s i n 4 x = - 16 x$

$\Rightarrow a = - 16 x$

$ω^{2} = 16 \Rightarrow ω = 4 = \frac{2 π}{T} \Rightarrow T = \frac{2 π}{4} = \frac{π}{2} s e c$

k = 2

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The potential energy of a particle of mass 4 kg in motion along the x-axis is given by $U = 4 (1 - c o s 4 x) J .$ The time period of the particle for small oscillation (sin = ) is $(\frac{π}{K}) s .$ The value of K is_______.

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The potential energy of a particle of mass 4 kg in motion along the x-axis is given by U=4(1−cos4x)J . The time period of the particle for small oscillation (sin = ) is (πK)s. The value of K is_______.

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The potential energy of a particle of mass 4 kg in motion along the x-axis is given by $U = 4 (1 - c o s 4 x) J .$ The time period of the particle for small oscillation (sin = ) is $(\frac{π}{K}) s .$ The value of K is_______.