A 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)
1. $ 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $
2. $ 2\pi \sqrt{\frac{v_1^2 - v_2^2}{x_1^2 + x_2^2}} $
3. $ 2\pi \sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}} $
4. $ 2\pi \sqrt{\frac{x_1^2 + x_2^2}{v_1^2 + v_2^2}} $
View Answer
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}} $.
A particle is executing a simple harmonic motion. Its maximum acceleration is $ \alpha $ and maximum velocity is $ \beta $. Then, its time period of vibration will be:
(2015 Re)
1. $ \frac{2\pi \beta}{\alpha} $
2. $ \frac{\beta^2}{\alpha^2} $
3. $ \frac{\alpha}{\beta} $
4. $ \frac{\beta^2}{\alpha} $
View Answer
Maximum acceleration is $ a_{\text{max}} = A\omega^2 = \alpha $ and maximum velocity is $ v_{\text{max}} = A\omega = \beta $. Dividing them, we get $ \omega = \frac{\alpha}{\beta} $. Thus, the time period is $ T = \frac{2\pi}{\omega} = \frac{2\pi \beta}{\alpha} $.
The phase difference between displacement and acceleration of a particle in a simple harmonic motion is: (2020)
1. $ \frac{3\pi}{2} \text{ rad} $
2. $ \frac{\pi}{2} \text{ rad} $
3. Zero
4. $ \pi \text{ rad} $
View Answer
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} $.
Average velocity of a particle executing SHM in one complete vibration is : (2019)
1. $ \frac{A\omega}{2} $
2. $ A\omega $
3. $ \frac{A\omega^2}{2} $
4. Zero
View Answer
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.
Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to paths of the two particles. The phase difference is: (2011 Mains)
1. $0$
2. $\frac{2\pi}{3}$
3. $\pi$
4. $\frac{\pi}{6}$
View Answer
Let displacement be $x = A\sin(\omega t + \phi)$. When they cross, $x = A/2$. $A/2 = A\sin(\phi) \Rightarrow \sin(\phi) = 1/2$. The two phases are $\pi/6$ and $5\pi/6$ (since moving in opposite directions). Phase difference $= 5\pi/6 - \pi/6 = 4\pi/6 = 2\pi/3$.
Out of the following functions representing motion of a particle which represents S.H.M.: (2011 Pre)n(i) $y = \sin \omega t – \cos \omega t$n(ii) $y = \sin^3 \omega t$n(iii) $y = 5\cos\left(\frac{3\pi}{4} – 3\omega t\right)$n(iv) $y = 1 + \omega t + \omega^2 t^2$
1. Only (i)
2. Only (iv) does not represent SHM
3. Only (i) and (iii)
4. Only (i) and (ii)
View Answer
(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.
The 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)
1. Simple harmonic motion of frequency $\omega/2\pi$
2. Simple harmonic motion of frequency $\omega/\pi$
3. Simple harmonic motion of frequency $3\omega/2\pi$
4. Non simple harmonic motion
View Answer
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 $f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.
Which one of the following equations of motion represents simple harmonic motion? (2009)nwhere $k$, $k_0$, $k_1$ and $a$ are all positive.
1. Acceleration $= -k(x)$
2. Acceleration $= k(x + a)$
3. Acceleration $= kx$
4. Acceleration $= -k_0x + k_1x^2$
View Answer
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 $\text{Acceleration} = -k(x)$.