The maximum displacement of a particle executing SHM from its mean position is 2 cm and its time period is 1 s. The equation of its displacement will be

Name: The maximum displacement of a particle executing SHM from its mean position is 2 cm and its time period is 1 s. The equation of its displacement will be
Uploaded: 2020-01-06
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Question

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Answer 1

Max. displacement = Amplitude = 2cm, thus it's of the form $x = 2 s i n ω t$ (assuming initial phase as 0)
Now, $ω = 2 π / T = 2 π / 1 = 2 π$
Thus, $x = 2 s i n 2 π t$

Answer 2

Max. displacement = Amplitude = 2cm, thus it's of the form

x = 2 s i n ω t

(assuming initial phase as 0)
Now,

ω = 2 π / T = 2 π / 1 = 2 π

Thus,

x = 2 s i n 2 π t

The maximum displacement of a particle executing SHM from its mean position is 2 cm and its time period is 1 s. The equation of its displacement will be

$x = 2 sin 4 π t$

$x = 2 sin 2 π t$

$x = sin 2 π t$

$x = 4 sin 2 π t$

Max. displacement = Amplitude = 2cm, thus it's of the form $x = 2 s i n ω t$ (assuming initial phase as 0)
Now, $ω = 2 π / T = 2 π / 1 = 2 π$
Thus, $x = 2 s i n 2 π t$

A particle starts its SHM from mean position at $t = 0$ . If its time period is $T$ and amplitude $A$ . The distance travelled by the particle in the time from $t = 0$ to $t = 5 T / 4$ is

The displacement-time graph of a particle executing SHM is shown in the figure. Then

A body executing SHM has its velocity $10 c m / s$ and $7 c m / s$ when its displacements from mean position are $3 c m$ and $4 c m$ respectively. The length of path is

The acceleration (a) of SHM at mean position is :

A particle is executing SHM with amplitude $A$ , time period $T$ , maximum acceleration $a_{0}$ and maximum velocity $v_{0}$ . It starts from mean position at $t = 0$ and at time $t$ it has the displacement $A / 2$ , acceleration $a$ and velocity $v$ , then

Revise with Concepts

Time Equations of SHM

The maximum displacement of a particle executing SHM from its mean position is 2 cm and its time period is 1 s. The equation of its displacement will be

x=2sin4πt

x=2sin2πt

x=sin2πt

x=4sin2πt

Max. displacement = Amplitude = 2cm, thus it's of the form x=2sinωt (assuming initial phase as 0) Now, ω=2π/T=2π/1=2π Thus, x=2sin2πt

A particle starts its SHM from mean position at t=0. If its time period is T and amplitude A. The distance travelled by the particle in the time from t=0 to t=5T/4 is

The displacement-time graph of a particle executing SHM is shown in the figure. Then

A body executing SHM has its velocity 10cm/s and 7cm/s when its displacements from mean position are 3cm and 4cm respectively. The length of path is

The acceleration (a) of SHM at mean position is :

A particle is executing SHM with amplitude A, time period T, maximum acceleration a0​ and maximum velocity v0​. It starts from mean position at t=0 and at time t it has the displacement A/2, acceleration a and velocity v, then

Revise with Concepts

Time Equations of SHM

$x = 2 sin 4 π t$

$x = 2 sin 2 π t$

$x = sin 2 π t$

$x = 4 sin 2 π t$

Max. displacement = Amplitude = 2cm, thus it's of the form $x = 2 s i n ω t$ (assuming initial phase as 0)
Now, $ω = 2 π / T = 2 π / 1 = 2 π$
Thus, $x = 2 s i n 2 π t$

A particle starts its SHM from mean position at $t = 0$ . If its time period is $T$ and amplitude $A$ . The distance travelled by the particle in the time from $t = 0$ to $t = 5 T / 4$ is

A body executing SHM has its velocity $10 c m / s$ and $7 c m / s$ when its displacements from mean position are $3 c m$ and $4 c m$ respectively. The length of path is

A particle is executing SHM with amplitude $A$ , time period $T$ , maximum acceleration $a_{0}$ and maximum velocity $v_{0}$ . It starts from mean position at $t = 0$ and at time $t$ it has the displacement $A / 2$ , acceleration $a$ and velocity $v$ , then