A particle in SHM is described by the displacement function- xt- A cos - t- - If the initial t-0 position of the particle is 1 cm and its initial velocity is - cm-sec - what is the amplitude-

The correct option is C √(2) cm x(t)=Acos(ω t+ϕ) v(t)=dx(t)dt = ω Asin(ω t+ϕ) Initial position= x(t=0)=Acosϕ=1cm Initial velocity= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Wave Equation

A particle in...

Question

A particle in SHM is described by the displacement function: $x (t) = A c o s (ω t + ϕ) .$ If the initial (t=0) position of the particle is 1 cm and its initial velocity is $π$ cm/sec , what is the amplitude?

\sqrt{2} c m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{7} c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{2} c m$
$x (t) = A c o s (ω t + ϕ)$
$⟹ v (t) = \frac{d x (t)}{d t}$
$= - ω A s i n (ω t + ϕ)$
Initial position= $x (t = 0) = A c o s ϕ = 1 c m$
Initial velocity= $v (t = 0) = - A ω s i n ϕ = ω c m / s$
Hence, $A = \sqrt{2} c m$

Suggest Corrections

Similar questions

Q. A particle in SHM is described by the displacement function:

x (t) = A c o s (ω t + ϕ) .

If the initial

(t = 0)

position of the particle is

1

cm and its initial velocity is

ω

cm/sec , what is the initial phase angle ?

Q. A particle in S.H.M is described by the displacement function

x (t) = a cos (ω t + θ)

. If the initial (t = 0) position of the particle is 1 cm and its initial velocity is

π

cm/s. The angular frequency of the particle is

π r a d / s

, then its amplitude

A particle in SHM is described by the displacement

equation x(t) = A cos ( $ω t + θ)$ . If the initial (t = 0)

position of the particle is 1 cm and its initial

velocity is $π$ cm/s, what is its amplitude?

The angular frequency of the particle is $π s^{- 1}$

Q. A particle executing SHM is described by the displacement function

x (t) = A c o s (ω t + ϕ),

If the initial at

(t = 0)

position of the particle is

1 c m,

its initial velocity is

π c m s^{- 1}

and its angular frequency is

π r a d s^{- 1},

then the amplitude of its motion is:

Q. The motion of a particle executing simple harmonic motion is described by the displacement function,

x (t) = A cos (ω t + ϕ)

.
If the initial

(t = 0)

position of the particle is 1 cm and its initial velocity is

ω

cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is

π s^{- 1}

. If instead of the cosine function we choose the sine function to describe the

S H M : x = B sin (ω t + α)

, what are the amplitude and initial phase of the particle with the above initial conditions.

Join BYJU'S Learning Program

A particle in SHM is described by the displacement function: x(t)=A cos(ωt+ϕ). If the initial (t=0) position of the particle is 1 cm and its initial velocity is π cm/sec , what is the amplitude?

The correct option is C √2cmx(t)=Acos(ωt+ϕ)⟹v(t)=dx(t)dt=−ωAsin(ωt+ϕ)Initial position=x(t=0)=Acosϕ=1cmInitial velocity=v(t=0)=−Aωsinϕ=ωcm/sHence, A=√2cm

A particle in SHM is described by the displacement function: $x (t) = A c o s (ω t + ϕ) .$ If the initial (t=0) position of the particle is 1 cm and its initial velocity is $π$ cm/sec , what is the amplitude?

The correct option is C $\sqrt{2} c m$
$x (t) = A c o s (ω t + ϕ)$
$⟹ v (t) = \frac{d x (t)}{d t}$
$= - ω A s i n (ω t + ϕ)$
Initial position= $x (t = 0) = A c o s ϕ = 1 c m$
Initial velocity= $v (t = 0) = - A ω s i n ϕ = ω c m / s$
Hence, $A = \sqrt{2} c m$