Question 71:
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 72:
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 73:
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 74:
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 $$\frac{2\pi}{T}t = \pi/6 \Rightarrow t = T/12$$.
Question 75:
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 76:
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$\omega^2 = 100 \implies \omega = 10 \text{ rad/s}$.
Question 77:
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 78:
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 79:
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 80:
moderateA simple harmonic oscillator has an amplitude $A$ and time period $T$. The time required by it to travel from $X = A$ to $X = A/2$ is:
(1992)
Using equation for SHM starting from extreme position: $x = A\cos(\omega t)$.\nSubstitute $x = A/2$: $A/2 = A\cos(2\pi t/T) \implies \cos(2\pi t/T) = 1/2$.\n$2\pi t/T = \pi/3 \implies t = T/6$.