Physics

Oscillations and Simple Harmonic Motion

136 Questions

Oscillations and simple harmonic motion focus on amplitude, damped vibrations, and force equations. These physics principles are essential for various engineering and civil services examinations. Review these problems to understand the core mechanics of oscillating bodies.

Damped harmonic oscillationSHM amplitudeForce equationsForced oscillationVibratory motion concepts

Oscillations and Simple Harmonic Motion Questions

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

The motion represented by equation $x=2\sin \omega t+3\sin^{2}\omega t$ is

  1. Periodic

  2. Oscillatory
  3. $SHM$

  4. Both (1) & (2)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation x = 2*sin(wt) + 3*sin^2(wt) can be written as 2*sin(wt) + 1.5*(1 - cos(2wt)). This is a sum of two periodic functions with different frequencies, so the motion is periodic and oscillatory, but not SHM because it is not a single sine/cosine term.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

A particle of mass $0.1 kg$ executes SHM under a force $F = (-10 x) N$. Speed of particle at mean position is $6 m/s$. Then amplitude of oscillations is 

  1. $0.6 m$
  2. $0.2 m$
  3. $0.4 m$
  4. $0.1 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} M=0.1\, Kg \ f=-10x \ \Rightarrow k=10 \ w=\sqrt { \frac { k }{ m }  }  \ =\sqrt { \frac { { 10 } }{ { 0.1 } }  } =10 \ { V _{ \max   } }=6m/s \ \Rightarrow A=\frac { 6 }{ { 10 } } =0.6m \ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Equation of SHM is $x=10\sin{10\pi t}$. Find the distance between the two points where speed is $50\pi$ $cm/sec$. $x$ is in cm and $t$ is in seconds

  1. $10cm$
  2. $20cm$
  3. $17.32cm$
  4. $8.66cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given x = 10 sin(10 pi t), the maximum velocity is omega times A = (10 pi)(10) = 100 pi cm/sec. The speed v at position x is given by omega times the square root of (A squared minus x squared). Setting v equal to 50 pi gives 50 pi = 10 pi times the square root of (100 minus x squared), which simplifies to 5 = square root of (100 minus x squared), giving x = plus or minus 5 times square root of 3 cm. The distance between these two points is 2 times 5 times square root of 3, which equals 10 times 1.732 = 17.32 cm.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Which of the following equation does not represent a simple harmonic motion:

  1. $y=a\sin\omega t$
  2. $y=b\cos\omega t$
  3. $y=a\sin\omega t+b\cos\omega t$
  4. $y= a\tan\omega t$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

SHM requires a restoring force proportional to displacement, resulting in a sine or cosine function of time. Tangent functions do not represent SHM.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Simple harmonic oscillation of a given system can be specified completely by stating its: 

  1. amplitude, frequency and initial phase.

  2. amplitude, frequency and wavelength

  3. frequency and wavelength.

  4. frequency, wavelength and initial phase.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Although waves consist of oscillation, there is no wavelength in a pure oscillation. It is there only in waves. 

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Simple harmonic motion (SHM) is a technical term used to describe a certain kind of idealized oscillation. Practically all the oscillations that one can see directly in the natural world are much more complicated than SHM. Why then do physicists make such a big deal out of studying SHM?

  1. It is the only kind of oscillation that can be described mathematically

  2. Any real oscillation can be analysed as a superposition (sum or integral) of SHMs with different frequencies

  3. Physics is concerned mainly with the unnatural world.

  4. Students are too stupid to appreciate the real world.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Simple harmonic motion provides a basis for the characterization of more complicated motions through the techniques of Fourier analysis. Here any waveform can be represented as closely as desired by the combination of a sufficiently large number of sinusoidal waves that form a harmonic series.   Fourier's theorem suggests that any periodic function can be represented as an algebraic sum of sine and cosine functions called a Fourier Series. 

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

The displacement of a particle is represented by the equation $y=sin^3(\omega t)$. The motion is 

  1. non-periodic

  2. periodic but not simple harmonic
  3. simple harmonic with period $\dfrac{2 \pi}{\omega}$
  4. simple harmonic with period $\dfrac{\pi}{\omega}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given the equation of displacement of the particle, $y={ sin }^{ 3 }\omega t$

We know $sin3\theta =3sin\theta -4{ sin }^{ 3 }\theta $
Hence, $y=\frac { (3sin\omega t-4sin3\omega t) }{ 4 } \ \Rightarrow 4\frac { dy }{ dt } =3\omega cos\omega t-4\times [3\omega cos3\omega t]\ \Rightarrow 4\times \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =-3{ \omega  }^{ 2 }sin\omega t+12\omega sin3\omega t\ \Rightarrow \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =\frac { -3{ \omega  }^{ 2 }sin\omega t+12\omega sin3\omega t }{ 4 } \ \Rightarrow \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } $ is not proportional to y. 
Hence, the motion is not SHM. 
As the expression is involving sine function, hence it will be periodic. 
Also ${ sin }^{ 3 }\omega t={ \left( sin\omega t \right)  }^{ 3 }\ ={ [sin(\omega t+2\pi )] }^{ 3 }\ ={ [sin(\omega t+2\pi /\omega )] }^{ 3 }$
Hence, $y={ sin }^{ 3 }\omega t$ represents a periodic motion with period $2\pi /\omega $.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

Which one of the following will take place when a watch based on oscillating spring is taken to a deep mine?

  1. It will indicate the same time on earth

  2. It will become fast

  3. It will become slow

  4. It will stop working
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A watch based on an oscillating spring (like a balance wheel watch) has a period T = 2*pi*sqrt(m/k). This period depends only on the mass and the spring constant, neither of which changes with gravity or depth. Thus, it keeps the same time.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

A simple harmonic oscillator of angular frequency 2 rad $s^{-1}$ is acted upon by an external force $F= sin tN.$ If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to 

  1. $ sin t+ \frac{1}{2}cos 2t$
  2. $ cos t -\frac{1}{2}sin2t$
  3. $sint -\frac{1}{2}sin2t$
  4. $sin t +\frac{1}{2}sin 2t$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The solution contains term proportional to $sin t$, $sin 2t$ & the only option consistent with initial conditions is (C)

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A transverse wave on a string has an amplitude of $02m$ and a frequency of $175Hz$. Consider a particle of the string at $x=0$. It begins with a displacement $y=0$ at $t=0$, according to equation $y=0.2\sin{(kx+\omega t)}$. How much time passes between the first two instant when this particle has a displacement of $y=0.1m$>

  1. $1.9ms$
  2. $3.9ms$
  3. $2.4ms$
  4. $0.5ms$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A harmonic oscillator vibrates with amplitude of 4 cm and performs 150 oscillations in one minute. If the initial phase is $45\circ$ and it starts moving away from the equation of motion is 

  1. $\displaystyle 0.04\, sin\, \left ( 5 \pi t\, +\, \frac{\pi}{4} \right )$
  2. $\displaystyle 0.04\, sin\, \left ( 5 \pi t\, -\, \frac{\pi}{4} \right )$
  3. $\displaystyle 0.04\, sin\, \left ( 4 \pi t\, +\, \frac{\pi}{4} \right )$
  4. $\displaystyle 0.04\, sin\, \left ( 4 \pi t\, -\, \frac{\pi}{4} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of motion of harmonic oscillator is $y=Asin(\omega t+\phi _0)$

It is given that $A=0.04m$
and $\phi _0=54^{\circ}=\dfrac{45}{180}\pi=\dfrac{\pi}{4}$
$\omega=2\pi\nu=2\pi\times \dfrac{150}{60}=5\pi$
Thus $y=0.04sin(5\pi t+\dfrac{\pi}{4})$
Hence the answer is option A.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A particle moves with simple harmonic motion in a straight line. In first $\tau s,$, after starting from rest it travels a distance $a$, and in next $\tau s$ it travels $2a$, in same direction, then:

  1. amplitude of motion is $4a$
  2. time period of oscillations is $6$,
  3. amplitude of motion is $3a$$\tau $
  4. time period of oscillations is $8$,$\tau $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For SHM starting from rest at x = A, x(t) = A cos(omega * t). Distance traveled in time tau is A - A cos(omega * tau) = a. In next tau, distance is A cos(omega * tau) - A cos(2 * omega * tau) = 2a. Solving these equations leads to the amplitude being 4a.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The displacement from the position of equilibrium of a point $4\ cm$ from a source of sinusoidal oscillations is half the amplitude at the moment $t=\dfrac{T}{6} (T$ is the time period$)$. Assume that the source was at mean position at $t=0$. The wavelength of the running wave is 

  1. $0.96\ m$
  2. $0.48\ m$
  3. $0.24\ m$
  4. $0.12\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Going by the data given to us, this wave is sinusoidal in nature and the wave equation takes the form of
$y = A\sin( \omega t - kx),$ as it is given that at $t = 0,$  the source is at mean position.
Here$,\ x = 4\ cm = 0.04\ m$
$y = A/2$
Amplitude $= A$
$t = \dfrac{T}{6}$
We know that $ \omega  = 2 \dfrac{ \pi }{T}$
$\Rightarrow \dfrac{A}{2} = A\sin((2  \pi  / T)(T/6) - 0.04k)$
$\sin((2 \pi  / T)(T/6) - 0.04k) = 1/2$
$\Rightarrow ((2  \pi / T)(T/6) - 0.04k) =  \pi / 6$
$k =  \pi  / 0.24$
wavelength $ \lambda = 2\pi  / k = 0.48\ m$