A particle executes linear simple harmonic motion with an amplitude of $ 3 \text{ cm} $. When the particle is at $ 2 \text{ cm} $ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is: (2017-Delhi)
Given $ |v| = |a| $, we have $ \omega \sqrt{A^2 - x^2} = \omega^2 x $. Substituting $ A = 3 $ and $ x = 2 $, we get $ \sqrt{3^2 - 2^2} = \omega (2) \implies \omega = \frac{\sqrt{5}}{2} $. The time period is $ T = \frac{2\pi}{\omega} = \frac{4\pi}{\sqrt{5}} $.
The displacement of a particle executing simple harmonic motion is given by $ y = A_0 + A \sin \omega t + B \cos \omega t $. Then the amplitude of its oscillation is given by : (2019)
The given equation can be written as $ y - A_0 = A \sin \omega t + B \cos \omega t $. This represents simple harmonic motion about a shifted mean position $ A_0 $ with an amplitude of $ \sqrt{A^2 + B^2} $.
A particle executing simple harmonic motion of amplitude $5 \text{ cm}$ has maximum speed of $31.4 \text{ cm/s}$. The frequency of its oscillation is: (2005)
$v_{\text{max}} = A\omega = A(2\pi f)$.\nSubstituting the given values, $31.4 = 5 \times 2 \times 3.14 \times f$.\nSolving this, we get $f = 1 \text{ Hz}$.
A body is executing simple harmonic motion. When the displacements from the mean position is $4 \text{ cm}$ and $5 \text{ cm}$ the corresponding velocities of the body is $10 \text{ cm/sec}$ and $8 \text{ cm/sec}$. Then the time period of the body is: (1991)
Using $v^2 = \omega^2(A^2 - x^2)$, we get $100 = \omega^2(A^2 - 16)$ and $64 = \omega^2(A^2 - 25)$.\nDividing them yields $A^2 = 41$. Substituting back gives $\omega^2 = 100/25 = 4 \implies \omega = 2$.\nTime period $T = 2\pi/\omega = \pi \text{ sec}$.
When an oscillator completes 100 oscillation its amplitude reduced to $\frac{1}{3}$ of initial value. What will be its amplitude, oscillation when it completes 200 oscillation? (2002)
Amplitude after $n$ oscillations is given by $A = A_0 e^{-k n}$. For $n = 100$, $A_{100} = A_0 e^{-100k} = A_0 \frac{1}{3}$. For $n = 200$, $A_{200} = A_0 e^{-200k} = A_0 (e^{-100k})^2 = A_0 (\frac{1}{3})^2 = A_0 \frac{1}{9}$.
A wave travelling in the +ve x-direction having displacement along y-direction as $1text{ m}$, wavelength $2\pitext{ m}$ and frequency of $1/\pitext{ Hz}$ is represented by: (2013)
Given amplitude $A = 1text{ m}$, wavelength $\lambda = 2\pitext{ m}$, and frequency $f = 1/\pitext{ Hz}$. Wave number $k = 2\pi/\lambda = 1$. Angular frequency $\omega = 2\pi f = 2$. The wave equation for +ve x-direction is $y = A\sin(kx - \omega t) = \sin(x - 2t)$.
The phase difference between two waves, represented by: $y_1 = 10^{-6} \sin{100t + (x/50) + 0.5} \text{m}$, $y_2 = 10^{-6} \cos{100t + (x/50)} \text{m}$ where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately: (2004)
Write $y_2$ as a sine function: $y_2 = 10^{-6} \sin(100t + x/50 + \pi/2)$. The phase difference is $\Delta\phi = \frac{\pi}{2} - 0.5 = 1.57 - 0.5 = 1.07 \text{ radians}$.
44. When a string is divided into three segments of length $l_1$, $l_2$ and $l_3$ the fundamental frequencies of these three segments are $v_1$, $v_2$ and $v_3$ respectively. The original fundamental frequency ($v$) of the string is: (2012 Pre)
The fundamental frequency is inversely proportional to the length of the string ($l \propto 1/v$). Since the total length $l = l_1 + l_2 + l_3$, substituting lengths with reciprocal frequencies gives $\frac{1}{v} = \frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}$.
49. Standing waves are produced in $10 \text{ m}$ long stretched string. If the string vibrates in 5 segments and wave velocity is $20 \text{ m/s}$, the frequency is: (1997)
The string vibrates in 5 segments, so the total length $L = 5 \times (\lambda/2)$. This gives $10 = 2.5 \lambda \implies \lambda = 4 \text{ m}$. Frequency $f = \frac{v}{\lambda} = \frac{20}{4} = 5 \text{ Hz}$.