Question 41:
easyThe differential equation of motion of a particle executing SHM is \(\frac{d^2y}{dt^2} + Ky = 0\) where \(K\) is positive constant. The time period of the oscillation is given by
Comparing with the standard equation \(\frac{d^2y}{dt^2} + \omega^2 y = 0\), we find \(\omega = \sqrt{K}\). Thus, the time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).
Question 42:
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 43:
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 44:
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 45:
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 46:
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 $\text{Acceleration} = -k(x)$.
Question 47:
easyA simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $a$ and time period $T$. The speed of pendulum at $x = a/2$ will be:
(2009)
Velocity in SHM is given by $v = \omega\sqrt{A^2 - x^2}$. Here $A = a$, $x = a/2$, and $\omega = \frac{2\pi}{T}$. So, $v = \frac{2\pi}{T}\sqrt{a^2 - \frac{a^2}{4}} = \frac{2\pi}{T} \times \frac{a\sqrt{3}}{2} = \frac{\pi a \sqrt{3}}{T}$.
Question 48:
easyA point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
(2008)
Velocity $v = \frac{dx}{dt} = a\omega\cos(\omega t + \pi/6)$. Maximum velocity is $a\omega$. Given $v = \frac{a\omega}{2}$, so $\cos(\omega t + \pi/6) = 1/2$. This gives $\omega t + \pi/6 = \pi/3 \Rightarrow \omega t = \pi/6$. Since $\omega = 2\pi/T$, we get $\frac{2\pi t}{T} = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$.
Question 49:
easyTwo Simple Harmonic Motions of angular frequency $100 \text{ rad s}^{-1}$ and $1000 \text{ rad s}^{-1}$ have the same displacement amplitude. The ratio of their maximum accelerations is:
(2008)
Maximum acceleration in SHM is given by $a_{\text{max}} = \omega^2 A$. Since amplitude $A$ is the same, $a_{\text{max}} \propto \omega^2$. The ratio is $a_1 / a_2 = (\omega_1 / \omega_2)^2 = (100 / 1000)^2 = (1/10)^2 = 1 : 100 = 1 : 10^2$.
Question 50:
easyA particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:
(2007)
Equation from equilibrium is $x = a\sin(\omega t)$. For $x = a/2$, $a/2 = a\sin(\omega t) \Rightarrow \sin(\omega t) = 1/2$. This gives $\omega t = \pi/6$. Substituting $\omega = 2\pi/T$, we get $$\frac{2\pi}{T}t = \pi/6 \Rightarrow t = T/12$$.