Tag: example of simple harmonic motion

Quizzes (0) Questions (154)

Questions Related to example of simple harmonic motion

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

A simple pendulum is released when $\theta = \pi/6$. The time period of oscillation is

  1. $\displaystyle 2\pi\sqrt{\frac{l}{g}}$
  2. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{293}{288}\right)$
  3. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{288}{293}\right)$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

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

A pendulum suspended from the ceiling of an elevator at rest has time period ${ T } _{ 1 }$. When the elevator moves up with an acceleration 'a' its time period of oscillation becomes ${ T } _{ 2 }$ when the elevator moves down with an acceleration 'a', its period of oscillation become ${ T } _{ 3 }$ then

  1. ${ T } _{ 1 }=\sqrt { { T } _{ 2 }{ T } _{ 3 } } $
  2. ${ T } _{ 1 }=\sqrt { T _{ 2 }{ ^{ 2 }T _{ 3 } }^{ 2 } } $
  3. ${ T } _{ 1 }=\dfrac { \sqrt { 2 } { T } _{ 2 }{ T } _{ 3 } }{ \sqrt { {T _{ 2 }}^{ 2 }+{T _{ 3 } }^{ 2 } } } $
  4. ${ T } _{ 1 }=\dfrac { { T } _{ 2 }{ T } _{ 3 } }{ \sqrt { {T _{ 2 }}^{ 2 }+{T _{ 3 } }^{ 2 } } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

In a conical pendulum, when the bob moves in a horizontal circle of radius r, with uniform speed V, the string of length L describe a cone of semi-vertical angle $\theta$. The tension  in the string is given by 

  1. $T = \dfrac{mgl}{(L^2 - r^2)}$
  2. $ T = \dfrac{\sqrt {L^2 - r^2}}{mgl}$
  3. $ T = \dfrac{mgL}{\sqrt {L^2 - r^2}}$
  4. $ T = \dfrac{mgL}{(L^2 - r^2)^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a conical pendulum, T*cos(theta) = mg. From the geometry, cos(theta) = h/L = sqrt(L^2 - r^2)/L. Therefore, T = mg/cos(theta) = mgL / sqrt(L^2 - r^2).

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.