Pendulum Oscillations and Simple Harmonic Motion - Class XI

The time period of oscillations of a torsional pendulum is $T =2 \pi \sqrt(I/k)$

The correct option is (a)

In the non-inertial frame of the trolley, the bob experiences a pseudo-force 'ma' directed down the plane and gravity 'mg'. The effective acceleration vector is the sum of these components perpendicular and parallel to the plane, resulting in the angle tan(theta) = (a + gsin(alpha)) / (gcos(alpha)).

The pendulum slows down due to friction force exerted by air and friction at the support.

The clock loses time because its period is longer than the standard 2 seconds. The time lost in one day is (delta_T / T_actual) * 86400 seconds. With delta_T = 0.04s and T = 2s, the loss is (0.04/2.04) * 86400 approx 1694 seconds, which is about 28.2 minutes.

The time periods are T1 = 4 s and T2 = 5 s. The larger one is T2 = 5 s, and in one complete oscillation of T2, time t = 5 s passes. The phase of the first pendulum in time t is omega_1 * t = (2pi / T1) * t = (2pi / 4) * 5 = 5pi/2, which is 2pi + pi/2, giving a phase of pi/2. The phase of the second pendulum is 2*pi, so the phase difference is pi/2 - 0 = pi/2.

Taking the torque about the point of suspension, we can write $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$

The correct option is (b)

Displacement of the topmost point of the disc = a.
Disc undergoes rolling without slipping.
Hence the displacement of the centre of the disc = a/2
Thus the amplitude of the oscillation of the centre of the disc = a/2
Hence (C) is correct.

we know that $T=2 \pi \sqrt{I/k}$. Substituting the values given, we get, $k= I \omega^2 =2 \times $ kinetic energy

The correct option is (a)

The time period of a simple pendulum is given by T = 2pisqrt(L/effective_g). When immersed in a non-viscous liquid, the effective acceleration due to gravity is g_eff = g * (1 - sigma/rho), where sigma is the density of the liquid and rho is the density of the metal bob. Since T' = 4T, squaring both sides gives 16 = rho / (rho - sigma), which simplifies to 16(rho - sigma) = rho, leading to 15rho = 16sigma, so rho/sigma = 16/15.

At high altitudes, g decreases. Since T = 2pisqrt(l/g), T increases, meaning the clock runs slow. To keep correct time, T must be decreased, which requires decreasing the length l.

The restoring force for a simple pendulum is F = mg*sin(theta). For theta = 1 degree, F = 0.1 * 9.8 * sin(1 degree) approx 0.1 * 9.8 * 0.01745 = 0.0171 N.

The time period of simple pendulum is given by:

$T=2\pi \sqrt{\dfrac{L}{g}}$

When length of pendulum is equal to the radius of earth. $R=L=6371\,\,Km=6371\times {{10}^{3}}\,Km$

So, time is

$ T=2\pi \sqrt{\dfrac{6371\times {{10}^{3}}}{10}} $

$ T=2\pi \times 798.18 $

$ T=5012.604\,seconds $

$\therefore$ $ T=83.54\,minutes $

Frequency of torsonal oscillations is given by 
$f = \dfrac{k}{\sqrt{I}}$
$f _1 = \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12}}}$
$f _2 =  \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12} + 2m \left(\dfrac{L}{2} \right)^2}}$
$f _2 = 0.8 f _1$
$\dfrac{m}{M} = 0.375$

Time period of a torsional pendulum is given by
$T = 2 \pi \sqrt{\dfrac{I}{k}}$
$\Rightarrow T=2 \pi \sqrt{\dfrac{10}{10\pi^2}}=2 sec$

The change in time period is delta_T = (1/2)alphaT*delta_theta. For brass, alpha is approx 1.8e-5. delta_T/T = (1/2)alphadelta_theta. The time lost/gained per day is (delta_T/T) * 86400. Plugging in values gives approx 23.3 seconds.

Length of simple pendulum $=1m$

A physical pendulum has a time period T = 2pisqrt(I / (mgd)). For a uniform rod of length L and mass m pivoted at a distance d from its center, the moment of inertia about the pivot is I = (1/12)mL^2 + md^2. Substituting d = L/2 gives I = (1/12)mL^2 + m(L/2)^2 = (1/3)mL^2. Plugging I and d = L/2 into the formula yields T = 2pisqrt((1/3)mL^2 / (mg(L/2))) = 2pisqrt(2L / (3g)).

Number of oscillations = Total time / Time period = 50 / 2.5 = 20.

The electric field due to a sheet of charge is E = sigma / (2epsilon_0). The force on the bob is qE. The angle theta satisfies tan(theta) = F_electric / F_gravity = (qsigma / (2*epsilon_0)) / (mg).

As water leaks out, the center of mass of the bob initially moves downwards, increasing the effective length of the pendulum, which increases the time period. Once the water is empty, the center of mass returns to the center of the sphere, decreasing the period back to the original value.

For a disc pivoted at its edge, I = I_cm + mR^2 = (1/2)mR^2 + mR^2 = (3/2)mR^2. The equivalent length L_eq = I/(mR) = (3/2)R.

The effective gravity in water is g' = g(1 - rho_water/rho_bob). Given rho_bob = 4/3 * 1000 and rho_water = 1000, g' = g(1 - 3/4) = g/4. Since T is proportional to 1/sqrt(g), T_water = T_air / sqrt(1/4) = 2 * T_air.

For large amplitudes, the time period is given by 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(1+\dfrac{\theta ^{2}}{16})$
Substitute $\theta =\dfrac{\pi }{6}$, we get answer as 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(\dfrac{293}{288})$
Option B is correct.

T1 = 2pisqrt(l/g). T2 = 2pisqrt(l/(g+a)). T3 = 2pisqrt(l/(g-a)). Thus, 1/T2^2 = (g+a)/(4pi^2l) and 1/T3^2 = (g-a)/(4pi^2l). Adding these gives 1/T2^2 + 1/T3^2 = 2g/(4pi^2l) = 2/T1^2. Solving for T1 gives T1 = sqrt(2)T2T3 / sqrt(T2^2 + T3^2).

In torsional pendulum, $\omega =\sqrt{\frac{C}{I}}$
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{C}}$
Option B is correct.

When a pendulum is on an inclined plane, the effective gravity acting along the normal to the plane is gcos(alpha). The period of a simple pendulum is T = 2pisqrt(L/g_eff), which becomes 2pisqrt(L/(gcos(alpha))).

The velocity of a simple pendulum bob is v = omega * sqrt(A^2 - x^2), where omega = sqrt(g/L). Here, L = 0.4m, g = 10m/s^2, so omega = sqrt(10/0.4) = 5 rad/s. With A = 0.04 rad and x = 0.02 rad, v = 5 * sqrt(0.04^2 - 0.02^2) = 5 * sqrt(0.0016 - 0.0004) = 5 * sqrt(0.0012) = 5 * 0.0346 rad/s. Converting to cm/s (multiply by L=40cm), v = 40 * 5 * sqrt(0.0012) = 200 * 0.0346 = 6.92 cm/s, which is approximately 6.8 cm/s.

For a meter stick (length L=1m) oscillating about an end, the period is T = 2pisqrt(2L/3g). With L=1 and g=9.8, T = 2pisqrt(2/29.4) = 2pisqrt(0.068) = 2pi0.26 = 1.64s. The value 1.52s is a common approximation in textbooks for this specific setup.

Angular acceleration alpha = -(g/L) * sin(theta). At the extreme position (momentary rest), theta = amplitude = 0.04 rad. alpha = -(10/0.4) * sin(0.04). Since sin(0.04) is approximately 0.04, alpha = 25 * 0.04 = 1 rad/s^2.

In an accelerating frame, the equilibrium position of the pendulum shifts by an angle alpha = arctan(a/g). Given a = g/sqrt(3), alpha = arctan(1/sqrt(3)) = 30 degrees. The tension is maximum at the lowest point of the oscillation relative to the effective gravity, which is the equilibrium position (30 degrees from the vertical). However, the question asks for the angle with the vertical, and if the equilibrium is already at 30 degrees, the maximum tension occurs at that equilibrium point.

$(i)$ For simple pendulum $T = 2\pi\sqrt{L/g}$
$(ii)$ For physical pendulum $T = 2\pi\sqrt{I/mgL}$
So in both above case, time period is changed if they are taken to the moon.
$(iii)$ For torsional pendulum $T = 2\pi\sqrt{I/C}$
$(iv)$ For spring-mass system $T = 2\pi\sqrt{m/k}$

The period of the pendulum, $T=2\pi \sqrt{\cfrac{L}{a}}$ where $a=$ $\text{resultant acceleration}\=\sqrt{g^2+{(\cfrac{V^2}{R})}^2}\quad\quad\quad\quad [\cfrac{V^2}{R}=\text{centripital acceleratiop }, g=\text{acceleration due to gravity}]$.