JEE Physics Oscillations MCQs Set 03

Answer: (d) \( 4\pi^2 m n^2 \)

Explanation:
1. Angular frequency \( \omega = 2\pi n \).
2. Force constant \( k = m\omega^2 = m(2\pi n)^2 \) = \( 4\pi^2 m n^2 \).

Teacher's Note:
1. The force constant does not depend on the amplitude.
2. Use \( k = m\omega^2 \) and \( \omega = 2\pi n \).

Answer: (a) \( \dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x \)

Explanation:
1. In S.H.M. the restoring force is \( F = -kx \).
2. By Newton's law \( m\,\dfrac{d^2x}{dt^2} = -kx \), so \( \dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x \).

Teacher's Note:
1. The acceleration is proportional to the displacement and opposite to it.
2. The minus sign shows that the force is restoring.

Answer: (a) \( 2\pi/\sqrt{k} \)

Explanation:
1. Comparing with \( \dfrac{d^2x}{dt^2} + \omega^2 x = 0 \) gives \( \omega^2 = k \).
2. \( T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{\sqrt{k}} \).

Teacher's Note:
1. Here k stands for the quantity that equals \(\omega^2\), not the spring constant.
2. Always compare the coefficient of x with \(\omega^2\).

Answer: (d) \( y = a\tan\omega t \)

Explanation:
1. Sine, cosine and any combination \( a\sin\omega t + b\cos\omega t = R\sin(\omega t + \phi) \) are S.H.M.
2. \( y = a\tan\omega t \) has \( \dfrac{d^2y}{dt^2} \) not proportional to \( -y \), and it is unbounded, so it is not S.H.M.

Teacher's Note:
1. S.H.M. is a single sinusoid of one frequency.
2. A tangent function is periodic but not harmonic.

Answer: (b) \( \pi/10 \)

Explanation:
1. Compare with \( y = A\sin(\omega t + \phi) \): \( \omega = 20 \) rad/s.
2. \( T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{20} = \dfrac{\pi}{10} \) s.

Teacher's Note:
1. The number multiplying t is the angular frequency.
2. T = 2\(\pi\)/\(\omega\).

Answer: (a) \( \pi/2 \)

Explanation:
1. With \( y = A\sin\theta \), the displacement is maximum when \( \sin\theta = 1 \).
2. That occurs at \( \theta = \pi/2 \).

Teacher's Note:
1. At the mean position the phase is 0 or \(\pi\) for a sine function.
2. The extreme positions are at \(\pi/2\) and \(3\pi/2\).

Answer: (b) \( 3000/2\pi \)

Explanation:
1. The angular frequency is \( \omega = 3000 \) rad/s.
2. \( f = \dfrac{\omega}{2\pi} = \dfrac{3000}{2\pi} \) Hz.

Teacher's Note:
1. 0.34 is the amplitude and 0.74 is the initial phase.
2. Frequency and angular frequency differ by a factor 2\(\pi\).

Answer: (a) always directed towards the equilibrium position

Explanation:
1. \( a = -\omega^2 x \), so the acceleration always points towards the mean position.
2. It is zero at the mean position and maximum at the extremes.

Teacher's Note:
1. The direction is always opposite to the displacement.
2. The acceleration changes sign as the particle crosses the mean position.

Answer: (a) four times the amplitude

Explanation:
1. In one period the particle goes from the mean to one extreme (A), back to the mean (A), to the other extreme (A) and back (A).
2. Total distance = 4A.

Teacher's Note:
1. Distance is different from displacement; the displacement in one period is zero.
2. 4A is the distance in one period.

Answer: (d) its displacement will be maximum

Explanation:
1. At \( \theta = \pi/2 \), \( y = A\sin(\pi/2) = A \): the displacement is maximum.
2. There the velocity is zero, and the acceleration and restoring force are maximum.

Teacher's Note:
1. At maximum displacement the velocity is zero and the force is largest.
2. At the mean position it is the other way round.

Answer: (c) 200 m/s

Explanation:
1. \( v_{max} = A\omega = 10 \times 20 \) = 200 m/s.

Teacher's Note:
1. Maximum velocity = amplitude \(\times\) angular frequency.
2. Read A and \(\omega\) directly from the equation.

Answer: (c) 60°

Explanation:
1. The constant term in the argument is \( \pi/3 \) rad.
2. \( \pi/3 \) rad = 60°.

Teacher's Note:
1. The phase constant is the phase at t = 0.
2. Convert radians to degrees with \(\pi\) rad = 180°.

Answer: (a) the particle is at \( x = a/2 \) and moving in the +X direction

Explanation:
1. \( x = a\sin(\pi/6) = a/2 \).
2. Velocity \( v = a\omega\cos(\pi/6) > 0 \), so the particle is moving in the +X direction.

Teacher's Note:
1. Displacement comes from the sine and velocity from the cosine of the phase.
2. The sign of the velocity gives the direction.

Answer: (c) \( \pi \)

Explanation:
1. Both pass the mean position together but with opposite velocities.
2. That means their displacements are always equal and opposite, which is a phase difference of \( \pi \).

Teacher's Note:
1. Opposite motion at the same instant means they are in antiphase.
2. Phase difference \(\pi\) means one is a mirror image of the other.

Answer: (a) 18 s

Explanation:
1. \( x = A\sin\left(\dfrac{2\pi t}{T}\right) \). With \( x = \dfrac{\sqrt{3}}{2}A \): \( \dfrac{2\pi\times 3}{T} = \dfrac{\pi}{3} \).
2. So \( T = 18 \) s.

Teacher's Note:
1. sin\(\theta\) = \(\sqrt{3}/2\) first happens at \(\theta\) = \(\pi/3\).
2. The particle starts from the mean position.

Answer: (b) 1 : 2

Explanation:
1. From \( \sin\omega t_1 = 1/2 \), \( \omega t_1 = \pi/6 \).
2. From \( \sin\omega t = 1 \), the time to the extreme is \( \omega t = \pi/2 \), so \( \omega t_2 = \pi/2 - \pi/6 = \pi/3 \).
3. \( t_1 : t_2 = \dfrac{\pi}{6} : \dfrac{\pi}{3} \) = 1 : 2.

Teacher's Note:
1. The particle slows down as it approaches the extreme, so the second interval is longer.
2. Use the phase angle, not the distance, to compare times.

Answer: (d) T/4

Explanation:
1. One full oscillation has four equal quarter stages: mean to extreme, extreme to mean, mean to the other extreme, and back.
2. So the time from mean to extreme is T/4.

Teacher's Note:
1. A quarter of the period corresponds to a quarter-cycle of phase, \(\pi/2\).
2. Time from extreme to the next mean position is also T/4.

Answer: (d) 1/2 s

Explanation:
1. \( x = A\sin(2\pi t/T) \) with \( x = A/2 \): \( \sin(2\pi t/6) = 1/2 \).
2. \( \dfrac{2\pi t}{6} = \dfrac{\pi}{6} \), so \( t = 0.5 \) s.

Teacher's Note:
1. Time to cover half the amplitude from the mean is T/12.
2. For T = 6 s this is 0.5 s.

Answer: (c) \( \pi \)

Explanation:
1. \( a = -\omega^2 x \): the acceleration is always opposite to the displacement.
2. So they are out of phase by \( \pi \).

Teacher's Note:
1. Velocity leads displacement by \(\pi/2\) and acceleration leads velocity by another \(\pi/2\).
2. The total phase difference between acceleration and displacement is \(\pi\).

Answer: (b) \( \pi/2 \)

Explanation:
1. If \( x = A\sin\omega t \), then \( v = A\omega\cos\omega t = A\omega\sin(\omega t + \pi/2) \).
2. The velocity leads the displacement by \( \pi/2 \).

Teacher's Note:
1. The velocity is zero at the extremes and maximum at the mean position.
2. So the two are a quarter cycle apart.

Answer: (d) \( v_0\sqrt{3}/2 \)

Explanation:
1. \( v = \omega\sqrt{A^2 - x^2} \) with \( x = A/2 \): \( v = \omega A\sqrt{1 - \tfrac{1}{4}} = \dfrac{\sqrt{3}}{2}\omega A \).
2. Since \( v_0 = \omega A \), \( v = \dfrac{\sqrt{3}}{2}v_0 \).

Teacher's Note:
1. "Position of rest" means the mean position, so half the amplitude is x = A/2.
2. v = \(\omega\)\(\sqrt{A^2 - x^2}\) is the standard velocity formula.

Answer: (a) \( a/2 \)

Explanation:
1. \( v = \omega\sqrt{a^2 - x^2} = \dfrac{\sqrt{3}}{2}\omega a \).
2. \( a^2 - x^2 = \dfrac{3}{4}a^2 \), so \( x^2 = \dfrac{a^2}{4} \) and \( x = a/2 \).

Teacher's Note:
1. The maximum velocity is \(\omega a\) at the mean position.
2. Square both sides to solve for x.

Answer: (a) 2

Explanation:
1. \( a = \omega^2 x \): \( \omega^2 = \dfrac{20}{5} = 4 \).
2. \( \omega = 2 \) rad/s.

Teacher's Note:
1. Use the magnitudes only.
2. Both a and x are in cm units, so no conversion is needed.

Answer: (a) 8

Explanation:
1. \( \omega = \dfrac{2\pi}{T} = 2 \) rad/s.
2. \( v = \omega\sqrt{A^2 - x^2} = 2\sqrt{25 - 9} = 2 \times 4 \) = 8 cm/s.

Teacher's Note:
1. The numbers 3, 4, 5 form a Pythagorean triple.
2. Use \(\omega = 2\pi/T\) first.

Answer: (a) 4

Explanation:
1. \( v_{max} = A\omega = 100 \) and \( a_{max} = A\omega^2 = 157 \).
2. Dividing: \( \omega = 1.57 \) rad/s.
3. \( T = \dfrac{2\pi}{\omega} = \dfrac{6.28}{1.57} \) = 4 s.

Teacher's Note:
1. The ratio of maximum acceleration to maximum velocity gives \(\omega\).
2. Note that 1.57 = \(\pi/2\).