Tag: free, forced and damped oscillations

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Questions Related to free, forced and damped oscillations

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation 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. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Amplitude A(t) = A0 * e^(-bt). Given A(t) = A0/2, e^(-bt) = 1/2. For 3t, A(3t) = A0 * (e^(-bt))^3 = A0 * (1/2)^3 = A0/8. Thus, n = 8.

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

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. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

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. $50 \,s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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

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

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

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

The frequency of vibration is less than the natural frequency in

  1. Forced vibrations

  2. Free vibration

  3. Damped vibrations

  4. All
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

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. $16$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

Periodic vibrations of decreasing amplitude are called

  1. Over Vibrations

  2. Critical Vibrations

  3. Damped Vibrations

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Damped vibrations occur when an oscillating system loses energy over time due to resistive forces like friction or air resistance, causing the amplitude to decrease.

Multiple choice physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In Melde's experiment, the frequency formula gives the relation between tension T and the number of loops n. For transverse vibration (fork at right angles), n1 * sqrt(T1) = 2 * n2 * sqrt(T2) for longitudinal vibration, or generally frequency is proportional to sqrt(T) for a fixed length. With 4 loops at 100 g in transverse mode, the fundamental frequency condition relates the tensions such that changing to 1 loop in longitudinal mode requires increasing the tension to 400 g. Since the initial tension is 100 g, an additional weight of 300 g is needed.

Multiple choice physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

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. $l/4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In perpendicular mode, f = (p/2L) * sqrt(T/m). In parallel mode, f = (p/L) * sqrt(T/m). To keep the same number of loops p and same frequency f, the length L must change. Setting (p/2L1) * sqrt(T/m) = (p/L2) * sqrt(T/m) results in L2 = 2 * L1.