Question 51:
easyThe phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:
(2007)
In SHM, velocity leads displacement by a phase of $\pi/2$, and acceleration leads velocity by a phase of $\pi/2$ (or $0.5\pi$). Thus, the phase difference between velocity and acceleration is $0.5\pi$.
Question 52:
easyOut of the following functions representing motion of a particle which represents S.H.M.:
(i) $y = \sin \omega t – \cos \omega t$
(ii) $y = \sin^3 \omega t$
(iii) $y = 5\cos\left(\frac{3\pi}{4} – 3\omega t\right)$
(iv) $y = 1 + \omega t + \omega^2 t^2$
(2011 Pre)
(i) Linear combination of sine and cosine represents SHM. (ii) $y = \sin^3\omega t$ is an oscillatory motion but not SHM. (iii) Simple cosine function with phase shift represents SHM. (iv) Non-oscillatory. Thus, only (i) and (iii) represent SHM.
Question 53:
easyA particle moves in a circle of radius $5 \text{ cm}$ with constant speed and time period $0.2\pi$. The acceleration of the particle is:
(2011 Pre)
Radius $r = 5 \text{ cm} = 0.05 \text{ m}$, Time period $T = 0.2\pi$. Angular velocity $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2\pi} = 10 \text{ rad/s}$. Centripetal acceleration $a = \omega^2 r = (10)^2 \times 0.05 = 100 \times 0.05 = 5 \text{ m/s}^2$.
Question 54:
easyWhich one of the following statements is true for the speed ‘$v$’ and the acceleration ‘$a$’ of a particle executing simple harmonic motion?
(2004)
In simple harmonic motion, speed is maximum at the mean position.\nAt this mean position, the displacement is zero, causing the restoring force and acceleration to be zero.
Question 55:
easyIf time of mean position from amplitude (extreme) position is $6\text{s}$. Then the frequency of S.H.M. will be:
(1998)
The time taken to travel from the extreme position to the mean position is $T/4$.\nThus, $T/4 = 6 \implies T = 24 \text{ s}$.\nFrequency $f = 1/T = 1/24 \approx 0.04 \text{ Hz}$.
Question 56:
easyA particle executes S.H.M. along x-axis. The force acting on it is given by:
(1994, 88)
For simple harmonic motion, the restoring force must be proportional to the negative of the displacement.\nThe equation $F = -Akx$ is the only one that satisfies the condition $F \propto -x$.
Question 57:
easyIf a simple harmonic oscillator has got a displacement of $0.02 \text{ m}$ and acceleration equal to $2 \text{ m/s}^2$ at any time, the angular frequency of the oscillator is equal to:
(1992)
Magnitude of acceleration in SHM is $|a| = \omega^2|x|$.\nSubstitute the values: $2 = \omega^2 \times 0.02$.\n$\omega^2 = 100 \implies \omega = 10 \text{ rad/s}$.