Home
Class 11
PHYSICS
The vertical motion of a ship at sea is ...

The vertical motion of a ship at sea is described by the equation `(d^(x))/(dt^(2))=-4x` , where x is the vertical height of the ship (in meter) above its mean position. If it oscillates through a height of 1 m

A

its maximum vertical speed will be `1 ms^(-1)`

B

its maximum vertical speed will be `2 ms^(-1)`

C

its greater vertical acceleration is `2 ms^(-2)`

D

its greater vertical acceleration is `1 ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B

(b) Comparing with `(d^(2)x)/(dt^(2))=-omega^(2)x`
we have , `omega=2 "rad"s^(-1)`
`v_("max")=omegaA=2ms^(-1)`
`a_("max")=omega^(2)A=4 ms^(-2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The motion of a particle is described by 9(d^(2)x)/(dt^(2))+25x=80 where x is displacement and t is time. Angular frequency of small oscillations of the particle

The roots of the equation x^(4)-2x^(2)+4=0 are the vertices of a :

The equation of SHM of a particle is given as 2(d^(2)x)/(dt^(2))+32x=0 where x is the displacement from the mean position. The period of its oscillation ( in seconds) is -

A body is projected horizontally from a point above the ground and motion of the body is described by the equation x=2t, y=5t^2 where x, and y are horizontal and vertical coordinates in metre after time t. The initial velocity of the body will be

The equation of motion of a particle executing SHM is ((d^2 x)/(dt^2))+kx=0 . The time period of the particle will be :

A body is projected horizontally from a point above the ground. The motion of the body is described by the equations x = 2 t and y= 5 t^2 where x and y are the horizontal and vertical displacements (in m) respectively at time t. What is the magnitude of the velocity of the body 0.2 second after it is projected?

The equation of motion of a particle of mass 1g is (d^(2)x)/(dt^(2)) + pi^(2)x = 0 , where x is displacement (in m) from mean position. The frequency of oscillation is (in Hz)

The equation of a particle executing SHM is (d^(2)x)/(dt^(2))=-omega^(2)x . Where omega=(2pi)/("time period") . The velocity of particle is maximum when it passes through mean position and its accleration is maximum at extremeposition. The displacement of particle is given by x=A sin(omegat+theta) where theta -initial phase of motion. A -Amplitude of motion and T-Time period The time period of pendulum is given by the equation T=2pisqrt((l)/(g)) . Here (d^(2)x)/(dt^(2)) is :