The displacement of two identical particles performing S.H.M are represented by equation x 1 = 6 sin ( 5 t + π /4 ) and x 2 = 4 cos ω t . The energies of both the particles will be same, if the va

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The displacement of two identical particles performing S.H.M are represented by equation x 1 ​ = 6 sin ( 5 t + π /4 ) and x 2 ​ = 4 cos ω t . The energies of both the particles will be same, if the value of ω is

TestHub · Accepted Answer

The total energy of a particle performing simple harmonic motion is given by $E = \frac{1}{2} m ω^{2} A^{2}$

For the first particle, the amplitude is $A_{1} = 6$ and the angular frequency is $ω_{1} = 5$ .

Its energy is $E_{1} = \frac{1}{2} m (5)^{2} (6)^{2} = \frac{1}{2} m (25) (36) = 450 m$

For the second particle, the amplitude is $A_{2} = 4$ and the angular frequency is $ω_{2} = ω$ .

Its energy is $E_{2} = \frac{1}{2} m (ω)^{2} (4)^{2} = 8 m ω^{2}$

Since the particles are identical, their masses are the same. Given that their energies are equal, we have $E_{1} = E_{2}$

$450 m = 8 m ω^{2}$

$ω^{2} = \frac{450}{8} = \frac{225}{4}$

$ω = \frac{15}{2} = 7.5$ unit

Answer: $7.5$ unit

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