Tag: free, forced and damped oscillations

Questions Related to free, forced and damped oscillations

The amplitude of a damped oscilator becomes one-half after $t$ second. If the amplitude becomes $\dfrac {1}{n}$ after $3t$, second, then $n$ is equal to

  1. $\dfrac {1}{8}$

  2. $8$

  3. $\dfrac {1}{4}$

  4. $4$


Correct Option: B

A system is executing forced harmonic resonant oscillations. The work done by the external driving force

  1. is equal to maximum K.E.

  2. is equal to maximum P.E.

  3. is equal to total energy

  4. is dissipated by damping forces


Correct Option: D
Explanation:

Generally, work done by external force goes to total energy of the system. But in forced oscillations, it is dissipated by damping forces.

Equation of motion for a particle performing damped harmonic oscillation is given as $x = e^{-1 t} cos (10 \pi t + \phi)$. The times when amplitude will half of the initial is :

  1. $27$

  2. $4$

  3. $1$

  4. $7$


Correct Option: D
Explanation:

$\dfrac{A _0}{2} = A _0 e^{-0.1t} \Rightarrow e^{-0.1t} = 2 \Rightarrow 0.1t = \ell n 2$
$t = \dfrac{\ell n 2}{0.1} = 10 \, \ell n2 \approx 6.93 \approx 7s$

A particle is performing damped oscillation with frequency $5Hz$. After every $10$ oscillations its amplitude becomes half. find time from beginning after which the amplitude becomes $\dfrac{1}{1000}$ of its initial amplitude:

  1. $10 \,s$

  2. $20 \,s$

  3. $25 \,s$

  4. $50 \,s$


Correct Option: B
Explanation:

$f = 5$
so $T = \dfrac{1}{5}$

$10T = \dfrac{10}{5} = 2$

$\dfrac{A _0}{1000} = A _0 \left(\dfrac{1}{2}\right)^{t/2}$

$(2)^{t/2} = 1000$

$\left(\dfrac{t}{2}\right) log 2 = 3$

$t = \dfrac{6}{log 2} \approx 20 s$

The frequency of vibration is less than the natural frequency in

  1. Forced vibrations

  2. Free vibration

  3. Damped vibrations

  4. All


Correct Option: C
Explanation:

It is our common experience that when a body is made to vibrate in a medium , the amplitude of the vibrating body continuously decreases with time and ultimately the body stops vibrating. This is called the damped vibrations.

A particle oscillating under a force $\bar{F} = - k \bar{x} - b \bar{v}$ is a (k and b are constants)

  1. simple harmonic oscillator

  2. linear oscillator

  3. damped oscillator

  4. forced oscillator


Correct Option: C
Explanation:

A particle oscillating under a force $\bar{F} = - k \bar{x} - b \bar{v}$ is damped oscillator. The first term $-k \bar{x}$ represents the restoring force and second term $-b \bar{v}$ represents the damping force.

In Melde's experiment, eight loops are formed with a tension of $0.75\space N$. If the tension is increased to four times then the number of loops produces will be

  1. $2$

  2. $4$

  3. $8$

  4. $16$


Correct Option: B
Explanation:

$Tp^2=constant$
$T _1P _1^2=T _2p _2^2$
$\dfrac{T _2}{T _1}=\dfrac{p _1^2}{p _2^2}$
$\Rightarrow \dfrac{p _1^2}{p _2^2}=4\Rightarrow p _2^2=\dfrac{p _1^2}{4}$
$\Rightarrow p _2^2=\dfrac{64}{4}=16$
$\Rightarrow p _2=4$

Periodic vibrations of decreasing amplitude are called

  1. Over Vibrations

  2. Critical Vibrations

  3. Damped Vibrations

  4. None of these


Correct Option: C

In Melde's experiment, when the tension is 100 g and the tuning fork vibrates at right angles to the direction of the string, 4 loops are produced. If now, the tuning fork is set to vibrate along the string, what additional weight will make the string vibrate in 1 loop? 

  1. 400 g

  2. 300 g

  3. 200 g

  4. 100 g


Correct Option: A

In Melde's experiment the position is changed from parallel to perpendicular. To get same number of loops, What should be the new length if original length is $l$? (Tension in the string is kept constant) 

  1. $2l$

  2. $l/2$

  3. $4l$

  4. $l/4$


Correct Option: A