JEE Physics Oscillations MCQs Set 16

Answer: (d) \( \vec F\cdot\vec a \)

Explanation:
1. \( \vec F = m\vec a \), so \( \vec F\cdot\vec a = ma^2\ge0 \).
2. \( \vec a \) and \( \vec F \) always point opposite to \( \vec r \), so \( \vec a\cdot\vec r \) and \( \vec F\cdot\vec r \) are never positive; \( \vec v\cdot\vec r \) changes sign during the cycle.

Teacher's Note:
1. Force and acceleration are parallel.
2. The restoring force is directed opposite to the displacement.

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

Explanation:
1. \( b\cos(\omega t-\phi) = b\sin(\omega t-\phi+\pi/2) \).
2. So the phase difference is pi/2.

Teacher's Note:
1. A cosine leads a sine by a quarter of a cycle.
2. The same phase constant phi appears in both and cancels.

Answer: (b) 4 s

Explanation:
1. Starting from rest means starting at an extreme: \( x = A\cos\omega t \).
2. \( \cos\omega t = 1/2 \Rightarrow \omega t = \pi/3 \), so \( t = T/6 = 4 \) s.

Teacher's Note:
1. Starting from rest implies the cosine form.
2. 60 degrees of the circle is one-sixth of the period.

Answer: (a) \( \dfrac{2A\omega^2}{\pi} \)

Explanation:
1. Starting at x = 0 with speed \( A\omega \), after half a period the particle is again at x = 0 with the velocity reversed: \( \Delta v = 2A\omega \).
2. \( a_{avg} = \dfrac{2A\omega}{T/2} = \dfrac{4A\omega}{T} = \dfrac{2A\omega^2}{\pi} \).

Teacher's Note:
1. Average acceleration is the change in velocity divided by the time taken.
2. The velocity at the end of half a cycle is the negative of the initial one.

Answer: (d) B

Explanation:
1. The constant A only shifts the mean position to y = A.
2. The oscillating part B sin(omega t) has amplitude B.

Teacher's Note:
1. Amplitude is measured from the mean position.
2. A constant added to y shifts the centre of the motion.

Answer: (c) \( 120^\circ \)

Explanation:
1. A displacement A/2 occurs at phases \( 30^\circ \) (moving outward) and \( 150^\circ \) (moving inward) for \( x = A\sin\theta \).
2. The two particles, moving in opposite directions at the same position, have phases \( 30^\circ \) and \( 150^\circ \): a difference of \( 120^\circ \).

Teacher's Note:
1. Use the reference circle: the same displacement occurs at two phases.
2. Opposite velocities mean the two phases are supplementary.

Answer: (b) A straight line with a negative slope and a positive intercept on the \( v^2 \)-axis

Explanation:
1. \( v^2 = \omega^2(A^2-x^2) \) and \( a^2 = \omega^4x^2 \), so \( v^2 = \omega^2A^2-\dfrac{a^2}{\omega^2} \).
2. This is a straight line in the \( (a^2, v^2) \) plane with negative slope and positive intercept \( \omega^2A^2 \).

Teacher's Note:
1. Eliminate x between the two expressions.
2. At a = 0 the speed is maximum, and at maximum a the speed is zero.

Answer: (a) maximum kinetic energy

Explanation:
1. At the mean position the displacement is zero, so the potential energy and the acceleration are zero (minimum).
2. The speed is maximum there, so the kinetic energy is maximum.

Teacher's Note:
1. Total energy is constant and shifts between kinetic and potential.
2. At the mean position all the energy is kinetic.

Answer: (d) \( A/\sqrt2 \)

Explanation:
1. \( \tfrac12k(A^2-x^2) = \tfrac12kx^2 \Rightarrow x^2 = A^2/2 \).
2. So \( x = A/\sqrt2 \).

Teacher's Note:
1. Each energy is half the total at that position.
2. Potential energy is (1/2) k x squared.

Answer: (b) The average kinetic energy over one time period equals the average potential energy over the same period

Explanation:
1. \( KE = \tfrac12kA^2\cos^2\omega t \) and \( PE = \tfrac12kA^2\sin^2\omega t \) both average to \( \tfrac14kA^2 \) over a full period.
2. They are equal at specific instants (x = A/over root 2) and not at all times, and over an arbitrary interval the averages need not match.

Teacher's Note:
1. Averages over a whole period (or half-period) are equal because sin squared and cos squared average to 1/2.
2. Over a short arbitrary interval the averages differ.

Answer: (a) \( \sqrt2\,A \)

Explanation:
1. At \( x = \tfrac{\sqrt3}{2}A \): \( PE = \tfrac12m\omega^2x^2 = \tfrac38m\omega^2A^2 \) and \( KE = \tfrac18m\omega^2A^2 \).
2. After the impulse \( KE' = \tfrac18+\tfrac12 = \tfrac58 \) (in units of \( m\omega^2A^2 \)), total \( = 1\times m\omega^2A^2 \).
3. \( \tfrac12m\omega^2A'^2 = m\omega^2A^2 \Rightarrow A' = \sqrt2A \).

Teacher's Note:
1. The impulse changes the kinetic energy but not the position, so the potential energy stays the same.
2. New total energy gives the new amplitude.

Answer: (d) \( \sqrt{k_2/k_1} \)

Explanation:
1. \( v_{max} = A\omega = A\sqrt{k/m} \). Equal masses and equal \( v_{max} \) give \( A_1\sqrt{k_1} = A_2\sqrt{k_2} \).
2. \( A_1/A_2 = \sqrt{k_2/k_1} \).

Teacher's Note:
1. Maximum speed is amplitude times the square root of k over m.
2. The stiffer spring needs the smaller amplitude.

Answer: (b) \( \dfrac{gT^2}{(2\pi)^2} \)

Explanation:
1. The extension at the mean position is \( x = mg/k \).
2. From \( T = 2\pi\sqrt{m/k} \): \( m/k = T^2/4\pi^2 \), so \( x = \dfrac{gT^2}{4\pi^2} \).

Teacher's Note:
1. The spring returns to its natural length when the ball is removed.
2. Replace m over k using the time period.

Answer: (d) become \( T/\sqrt2 \)

Explanation:
1. Cutting the spring in half doubles the spring constant: \( k' = 2k \).
2. \( T' = 2\pi\sqrt{m/2k} = T/\sqrt2 \).

Teacher's Note:
1. Spring constant is inversely proportional to the length of the spring.
2. Half the length means twice the stiffness.

Answer: (b) E

Explanation:
1. With the same spring, \( E = \tfrac12kA^2 \) depends only on k and A, both unchanged.
2. So the energy remains E (the angular frequency rises, but m omega squared stays equal to k).

Teacher's Note:
1. Energy of a spring oscillator is (1/2) k A squared.
2. m omega squared is just k.

Answer: (a) \( 49/81 \)

Explanation:
1. \( T_1/T_2 = 7/9 \) (the first has the shorter period).
2. \( \ell\propto T^2 \Rightarrow \ell_1/\ell_2 = 49/81 \).

Teacher's Note:
1. More oscillations in the same time means a shorter period.
2. Length is proportional to the square of the period.

Answer: (b) 0%

Explanation:
1. For small amplitudes the period of a simple pendulum does not depend on the amplitude.
2. So the percentage change is 0%.

Teacher's Note:
1. Isochronism holds for small oscillations.
2. Only length and g determine the period.

Answer: (a) \( mg\left[1+\left(\dfrac{A}{\ell}\right)^2\right] \)

Explanation:
1. At the lowest point \( T = mg+\dfrac{mv^2}{\ell} \).
2. For small amplitude \( \theta_0 = A/\ell \) and \( v^2 = 2g\ell(1-\cos\theta_0)\approx g\ell\theta_0^2 = \dfrac{gA^2}{\ell} \).
3. \( T_{max} = mg\left[1+\dfrac{A^2}{\ell^2}\right] \).

Teacher's Note:
1. Maximum tension occurs at the bottom of the swing.
2. Use small-angle approximation for the speed.

Answer: (c) \( \sqrt2f_0 \)

Explanation:
1. For a rod pivoted at one end, \( T = 2\pi\sqrt{\dfrac{2L}{3g}} \), so \( f\propto1/\sqrt L \).
2. With half the length, \( f' = \sqrt2f_0 \).

Teacher's Note:
1. A uniform rod pivoted at the end behaves like a simple pendulum of length 2L/3.
2. Halving the length multiplies the frequency by root two.

Answer: (b) a linear SHM along a straight line inclined equally to the straight lines of the component motions

Explanation:
1. With \( x = A\sin\omega t \) and \( y = A\sin\omega t \), we get \( y = x \).
2. The path is a straight line at 45 degrees to both axes, traversed with SHM.

Teacher's Note:
1. Equal phase gives a straight line; a quarter-cycle phase gives a circle or ellipse.
2. The amplitude along the line is A root two.

Answer: (a) SHM with amplitude \( \sqrt{C^2+D^2} \)

Explanation:
1. \( C\sin\omega t+D\cos\omega t = R\sin(\omega t+\phi) \) with \( R = \sqrt{C^2+D^2} \).
2. This is simple harmonic motion with that amplitude.

Teacher's Note:
1. Combine a sine and a cosine of the same frequency into one sinusoid.
2. The two terms are quarter-cycle apart so their amplitudes add in quadrature.

Answer: (d) 10 m

Explanation:
1. \( R = 5\sqrt{1^2+(\sqrt3)^2} = 5\times2 = 10 \) m.

Teacher's Note:
1. Combine the two terms into a single sine.
2. The amplitude is the square root of the sum of the squared coefficients.

Answer: (c) on a circle

Explanation:
1. \( x^2+y^2 = A^2 \), so the path is a circle of radius A.
2. It is a uniform circular motion, not SHM, and the acceleration is not constant (it changes direction).

Teacher's Note:
1. Square and add the two coordinates.
2. Two perpendicular SHMs a quarter-cycle apart with equal amplitudes give a circle.

Answer: (a) \( 1/9 \) of the initial value

Explanation:
1. The amplitude decays exponentially, so it falls by the same factor in each equal number of oscillations.
2. After 100 oscillations \( A_0/3 \); after the next 100, \( (A_0/3)/3 = A_0/9 \).

Teacher's Note:
1. Exponential decay multiplies the amplitude by a fixed factor in equal time intervals.
2. Square the factor for double the time.

Answer: (d) Amplitudes 30/4 and 10/4; frequencies 1/2 and 3/2

Explanation:
1. \( \sin^3\theta = \dfrac{3\sin\theta-\sin3\theta}{4} \), so \( x = \dfrac{30}{4}\sin\pi t-\dfrac{10}{4}\sin3\pi t \).
2. The terms have \( \omega = \pi \) and \( 3\pi \), i.e. frequencies 1/2 Hz and 3/2 Hz, with amplitudes 30/4 and 10/4.

Teacher's Note:
1. Use the triple-angle identity for sin cubed.
2. The amplitude pairs with the matching frequency in the order written.

Answer: (c) \( 6\sqrt2 \)

Explanation:
1. Sine components: \( 2+10\cos37^\circ-4 = 2+8-4 = 6 \).
2. Cosine components: \( 10\sin37^\circ-12 = 6-12 = -6 \).
3. Resultant amplitude \( = \sqrt{6^2+6^2} = 6\sqrt2 \).

Teacher's Note:
1. Resolve every term into sin and cos components of the same argument.
2. Add the two perpendicular components in quadrature.

Answer: (b) \( \omega = \dfrac{\omega_1\omega_2}{\sqrt{\omega_1^2+\omega_2^2}} \)

Explanation:
1. \( k_1 = m\omega_1^2 \) and \( k_2 = m\omega_2^2 \). In series \( \dfrac1k = \dfrac1{k_1}+\dfrac1{k_2} \).
2. \( \dfrac1{\omega^2} = \dfrac1{\omega_1^2}+\dfrac1{\omega_2^2} \Rightarrow \omega = \dfrac{\omega_1\omega_2}{\sqrt{\omega_1^2+\omega_2^2}} \).

Teacher's Note:
1. Springs in series: add the reciprocals of the spring constants.
2. The mass is the same so omega squared is proportional to k.

Answer: (a) 2f

Explanation:
1. \( KE = \tfrac12mA^2\omega^2\cos^2\omega t = \tfrac14mA^2\omega^2(1+\cos2\omega t) \).
2. The oscillating part has angular frequency 2 omega, i.e. frequency 2f.

Teacher's Note:
1. Kinetic energy depends on the square of the velocity.
2. It reaches a maximum twice in every period.

Answer: (c) 10

Explanation:
1. \( x_A = 3-2\cos2\pi t \), \( y_B = 4-2\sin2\pi t \).
2. \( d^2 = (3-2\cos)^2+(4-2\sin)^2 = 29-12\cos2\pi t-16\sin2\pi t = 29-20\cos(2\pi t-\phi) \).
3. So \( s_1 = \sqrt{49} = 7 \) cm and \( s_2 = \sqrt9 = 3 \) cm, giving 10.

Teacher's Note:
1. Write both coordinates as functions of time and form the squared distance.
2. The combination of cos and sin has amplitude 20.