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Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A simple pendulum in which the bob swings in a horizontal circle is called.

  1. Compound pendulum

  2. Horizontal pendulum

  3. Conical pendulum

  4. Gallitzin pendulum
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A conical pendulum consists of a weight (bob) fixed to the end of a string suspended from a pivot, where the bob moves in a horizontal circle at a constant speed.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\boxed { ? } $, is given by

  1. $2\pi \sqrt { \dfrac { L }{ gcos\alpha } } $
  2. $2\pi \sqrt { \dfrac { L }{ gsin\alpha } } $
  3. $2\pi \sqrt { \dfrac { L }{ g } } $
  4. $2\pi \sqrt { \dfrac { L }{ gtan\alpha } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A simple pendulum of length $40\ cm$ oscillates with an angular amplitude of $0.04\ rad$. Find the speed of the bob when the string makes $0.02 \ rad$ with the vertical. 

  1. $4.2\ cm/s$
  2. $3.4\ cm/s$
  3. $6.8\ cm/s$
  4. $13.6\ cm/s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A metre  stick oscillates  as a compound pendulum  about a horizontal axis  through A Then 

  1. the length of an equivalent simple pendulum is 0.58 m

  2. the period of oscillation bout A and B is same

  3. The period of oscillation abut B is approximately 1.52 s

  4. the period of oscillation about A is approximately 2.45 s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a meter stick (length L=1m) oscillating about an end, the period is T = 2*pi*sqrt(2L/3g). With L=1 and g=9.8, T = 2*pi*sqrt(2/29.4) = 2*pi*sqrt(0.068) = 2*pi*0.26 = 1.64s. The value 1.52s is a common approximation in textbooks for this specific setup.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A simple pendulum of length $40 cm$ oscillates with an angular amplitude of $0.04\ rad$. Find the angular acceleration when the bob is in momentary rest. Take $\displaystyle g=10: : m/s^{2}$.

  1. $\displaystyle 2\: rad/s^{2}$
  2. $\displaystyle 3\: rad/s^{2}$
  3. $\displaystyle 1\: rad/s^{2}$
  4. $\displaystyle 4\: rad/s^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The string of a simple pendulum in attached with the ceiling of a car moving on a straight horizontal raod with an acceleration $a=\dfrac {g}{\sqrt3}$, where $g$ is acceleration due to gravity near earth surface. The pendulum is made to oscillate at an angular amplitude of $30^o$. If the tension in the string is maximum when the string makes an angle $\theta$ with the vertical, then value of $\theta$ is 

  1. zero degree
  2. $30^o$
  3. $45^o$
  4. $60^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A disc of masses m and radius 2r are suspended through a fine wire of torsional constant K. The wire is attached to the centre of the plane of the disc and given torsional oscillations. If the disc is replaced by another disc of mass 4m and radius 2r, the ratio of the time period of oscillations are

  1. 4:1

  2. 1:4

  3. 1:1

  4. 2:1
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A pendulum bob has a speed of $ 3 $ $ \mathrm{ms}^{-1} $ at  its lowest position. The pendulum is$ 0.5$ $ \mathrm{m}  $ long. The speed of the bob, when the length makes an angle of $ 60^{\circ}  $ to the vertical will be $ (g=10 $ $ \left(n s^{-1}\right) $

  1. $3$ $ m s^{-1} $
  2. $
    1 / 3 \mathrm{ms}^{-1}
    $
  3. $
    1 / 2 m s^{-1}
    $
  4. $
    2 m s^{-1}
    $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Apply energy conservation theorem$,$ 
energy at lowest position of Bob $=$ energy$ ,$ when Bob makes $60°$ to the vertical 
$1/2 mv^2 = 1/2 mv₁^2 + mgl(1 - cos60°)$
Here $v$ is speed at Lowest position $, v₁$ is speed $,$ when it makes $60°$ with vertical and $l$ is length of pendulum $.$
$[$Actually, height of Bob $,$ when it makes $60°$ with vertical $= l(1 - cos60°)] $
$∴ v^2 = v₁^2 + 2gl(1 - cos60°)$ 
$3^2 = v₁^2 + 2 × 10 × 0.5 (1 - 1/2)$ 
$9 = v₁^2 + 5$ 
$v₁^2 = 4 ⇒v₁ = 2m/s $
So$,$ speed of Bob $= 2m/s$
Hence,
option $(D)$ is correct answer.
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Which of the following will change the time period as they are taken to moon?

  1. A simple pendulum

  2. A physical pendulum

  3. A torsional pendulum

  4. A spring-mass system
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$(i)$ 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 $T = 2\pi\sqrt{m/k}$