Tag: oscillations

Quizzes (16) Questions (206)

Questions Related to oscillations

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

A simple pendulum with a metal bob has a time period $T$. Now the bob is immersed in a liquid which is non viscous. This time the time period is $4T$. The the ratio of densities of metal bpob and that of the liquid is

  1. $15:16$
  2. $16:15$
  3. $1:16$
  4. $16:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of a simple pendulum is given by T = 2*pi*sqrt(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 15*rho = 16*sigma, so rho/sigma = 16/15.

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

A pendulum clock keeping correct time is taken to high altitudes,

  1. it will keep correct time

  2. its length should be increased to keep correct time

  3. its length should be decreased to keep correct time

  4. it cannot keep correct time even if the length is
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

Restoring force on the bob of a simple pendulum of mass $100\ gm$ when its amplitude is ${ 1 }^{ 0 } $ is 

  1. $0.017\ N$
  2. $1.7\ N$
  3. $0.17\ N$
  4. $0.034\ N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

Simple pendulum of large length is made equal to the radius of earth. Its period of oscillation will be then?

  1. 83.5 minutes

  2. 59.8 minutes

  3. 42.3 minutes
    1. 15 minutes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 $

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

If the length of a clock pendulum increase by $0.2\%$ due to atmospheric temperature rise, then the loss in time of clock per day is 

  1. $86.4$s
  2. $43.2$s
  3. $72.5$s
  4. $32.5$s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$T=2\pi \sqrt{\dfrac{l}{g}}$

$\dfrac{\Delta T}{T}\times 100 = \dfrac{1}{2}\dfrac{\Delta l}{l}\times 100 $

$\dfrac{\Delta T}{24 \times 3600}\times 100 = \dfrac{1}{2}\dfrac{0.2}{100}\times 100 $

$\Delta T=86.4s$

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

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are
attached at distance $'L/2'$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio m/M is close to :

  1. 0.175

  2. 0.375

  3. 0.575

  4. 0.775
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Find the time period of oscillations of a torsional pendulum, if the torsional constant of the wire is K = 10$\pi^2$J/rad. The moment of inertia of rigid body is 10 Kg m$^2$ about the axis of rotation.

  1. 2 sec

  2. 4 sec

  3. 16 sec

  4. 8 sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

A clock which has a pendulum made of brass keep correct time at ${30^0}C$? How many seconds it will gain or lose in day if the temperature falls to ${0^0}C$.

  1. It will lose $23.32$ sec per day
  2. It will gain $23.32$ sec per day
  3. It will lose $50$ sec per day
  4. It will gain $50$ sec per day
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

A simple pendulum of length $1$m has a bob of mass $100$g. It is displaced through an angle of $60^o$ from the vertical and then released . Find out K.E. of bob when it passes through mean position.

  1. $0.12$J
  2. $0.24$J
  3. $0.36$J
  4. $0.55$J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Length of simple pendulum $=1m$

Mass $=1w\;gm=0.1kg$
It is displaced through as angle of $60$ for vertical 
Height of pendulum at starting position $=$ length $(1-ws\;60)$
                                                                   $=1\left( 1-0.5\right)$
                                                                   $=0.5m$
Potential energy $=mgh=0.1\times 10\times 0.5 =0.55$
when it is released and it reaches mean position its potential energy at starting point is converted to kinetic energy.
so K.E. of bob at mean position $=0.55.$
Hence, the answer is $0.55.$