Question 61:
moderateIf \( x = 2\sin\left(\frac{\pi}{2}t\right) \) represents the motion of a particle executing SHM, the maximum speed of the particle in \( \text{m s}^{-1} \) is (All parameters are in SI units)
Comparing the given equation with the standard SHM equation \( x = A\sin(\omega t) \), we get \( A = 2\text{ m} \) and \( \omega = \frac{\pi}{2}\text{ rad/s} \). The maximum speed is \( v_{\max} = A\omega = 2 \times \frac{\pi}{2} = \pi\text{ m/s} \).
Question 62:
easyAverage velocity of a particle executing SHM in one complete vibration is :
(2019)
In one complete vibration, the particle returns to its starting point, so the net displacement is zero. Since average velocity is total displacement divided by total time, it is zero.
Question 63:
easyWhen two displacements represented by $ y_1 = a \sin(\omega t) $ and $ y_2 = b \cos(\omega t) $ are superimposed, the motion is:
(2015)
The resultant displacement is $ y = y_1 + y_2 = a \sin(\omega t) + b \cos(\omega t) $. This equation represents a single simple harmonic motion with a resultant amplitude of $ R = \sqrt{a^2 + b^2} $.
Question 64:
moderateA particle is executing S.H.M. along a straight line. Its velocities at distances $ x_1 $ and $ x_2 $ from the mean position are $ v_1 $ and $ v_2 $, respectively. Its time period is:
(2015)
Velocity in SHM is $ v^2 = \omega^2(A^2 - x^2) $. So, $ v_1^2 = \omega^2(A^2 - x_1^2) $ and $ v_2^2 = \omega^2(A^2 - x_2^2) $. Subtracting these equations gives $ v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2) $, yielding $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $$.
Question 65:
easyThe phase difference between displacement and acceleration of a particle in a simple harmonic motion is:
(2020)
Displacement is given by $ y = A \sin(\omega t) $ and acceleration is $ a = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi) $. Thus, the phase difference is $ \pi \text{ rad} $.
Question 66:
easyIdentify the function which represents a periodic motion.
(2020-Covid)
The function $ \sin \omega t + \cos \omega t $ represents a superposition of two simple harmonic motions, making it periodic. The other given functions do not repeat their values over equal intervals of time.
Question 67:
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 68:
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 69:
moderateThe displacement of a particle along the x-axis is given by $x = a\sin^2\omega t$. The motion of the particle corresponds to:
(2010 Pre)
Equation is $x = a\sin^2\omega t = \frac{a}{2}(1 - \cos 2\omega t)$. This represents SHM about the mean position $x = a/2$. The angular frequency is $2\omega$. The frequency is $f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.
Question 70:
easyWhich one of the following equations of motion represents simple harmonic motion?
where $k$, $k_0$, $k_1$ and $a$ are all positive.
(2009)
For simple harmonic motion, the acceleration must be directly proportional and opposite in direction to the displacement. Thus, $a \propto -x$, which matches the equation $\text{Acceleration} = -k(x)$.