The maximum velocity and the maximum acceleration of a body executing simple harmonic oscillator are 2 m/s and 4 m/s2 and Then angular velocity will be A. 3 rad/sec B. 0.5 rad/sec C. 1 rad/sec D. 2 rad/sec

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The correct option is (D)

The maximum acceleration of a body executing simple harmonic motion of amplitude A and angular velocity ω is given by

a_max = ω²A

The maximum velocity of a body executing simple harmonic motion of amplitude A and angular velocity ω is given by

v_max = ωA

Taking the ratio of maximum acceleration to maximum velocity, we get

\(\frac{a_{max}}{v_{max}} = \frac{\omega^2A}{\omega A} = \omega\)

Hence, the angular velocity of the body is the ratio of maximum acceleration to maximum velocity

⇒ ω = \(\frac{a_{max}}{v_{max}}\)

Given in question,

Maximum acceleration of the body, a_max = 4 m/s²

Maximum velocity of the body, v_max = 2 m/s

Therefore, the angular velocity of the body is given by

ω = \(\frac{a_{max}}{v_{max}}\) = \(\frac{4}{2}\) = 2 rad/sec

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The maximum velocity and the maximum acceleration of a body executing simple harmonic oscillator are 2 m/s and 4 m/s2 and Then angular velocity will be A. 3 rad/sec B. 0.5 rad/sec C. 1 rad/sec D. 2 rad/sec

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Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.
Write any two features of nuclear forces.

3.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

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The maximum velocity and the maximum acceleration of a body executing simple harmonic oscillator are 2 m/s and 4 m/s2 and Then angular velocity will be A. 3 rad/sec B. 0.5 rad/sec C. 1 rad/sec D. 2 rad/sec

CBSE CLASS XII Related Questions

1.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2. Write any two features of nuclear forces.

3.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.
Write any two features of nuclear forces.

3.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: