Question

If a particle of mass 50 g executes S.H.M along a path length of 20 cm at the rate of 600
oscillations per minute, find its energy. Also find the potential energy and kinetic energy at a
point midway between mean and extreme positions.
(Ans: (i) The total energy of the particle is 0.987 J.
(ii) The potential energy of the particle is 0.247 J.
(ii) The kinetic energy of the particle is 0.741 J.)​

Answer 1

Answer:

Part a)

Total energy = 0.987 J

Part b)

Potential energy = 0.247 J

Part c)

Kinetic energy = 0.741 J

Explanation:

As we know that number of oscillation per minute is given as

$f = \frac{600}{60}$

$f = 10 Hz$

now we have

$\omega = 2\pi f$

$\omega = 2\pi(10)$

Amplitude of SHM is given as

$2A = 20 cm$

$A = 10 cm$

Part a)

Total energy of SHM is given as

$E = \frac{1}{2}m\omega^2 A^2$

$E = \frac{1}{2}(0.050)(20\pi)^2(0.10)^2$

$E = 0.987 J$

Part b)

Now when $x = \frac{A}{2}$

$U = \frac{1}{2}m\omega^2(\frac{A}{2})^2$

$U = \frac{1}{8} m\omega^2 A^2$

$U = \frac{0.987}{4}$

$U = 0.247$

Part c)

As we know that sum of kinetic energy and potential energy is total energy of SHM

So we will have

$KE + U = E$

$KE = E - U$

$KE = 0.987 - 0.247$

$KE = 0.741 J$

#Learn

Topic : SHM energy

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