Advertisements
Advertisements
Question
The displacement of a particle varies with time according to the relation y = a sin ωt + b cos ωt.
Options
The motion is oscillatory but not S.H.M.
The motion is S.H.M. with amplitude a + b.
The motion is S.H.M. with amplitude a2 + b2.
The motion is S.H.M. with amplitude `sqrt(a^2 + b^2)`.
Solution
The motion is S.H.M. with amplitude `sqrt(a^2 + b^2)`.
Explanation:
According to the question, the displacement
y = a sin ωt + b cos ωt
Let a = A sin `phi` and b = A cos `phi`
Now, a2 + b2 = A2 sin2 `phi` + A2 cos2 `phi`
= A2
⇒ A = `sqrt(a^2 + b^2)`
y = A sin `phi` sin ωt + A cos `phi` cos ωt
= A sin (ωt + `phi`)
`(dy)/(dt) = Aω cos (ωt + phi)`
`(d^2y)/(dt^2) = - Aω^2 sin(ωt + phi)`
= `- Ayω^2`
= `(- Aω^2)y`
⇒ `(d^2y)/(dt^2) ∝ (-y)`
Hence, it is an equation of SHM with amplitude A = `sqrt(a^2 + b^2)`.
APPEARS IN
RELATED QUESTIONS
The average displacement over a period of S.H.M. is ______.
(A = amplitude of S.H.M.)
A particle executing simple harmonic motion comes to rest at the extreme positions. Is the resultant force on the particle zero at these positions according to Newton's first law?
In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.
It is proposed to move a particle in simple harmonic motion on a rough horizontal surface by applying an external force along the line of motion. Sketch the graph of the applied force against the position of the particle. Note that the applied force has two values for a given position depending on whether the particle is moving in positive or negative direction.
A pendulum clock keeping correct time is taken to high altitudes,
A small block oscillates back and forth on a smooth concave surface of radius R ib Figure . Find the time period of small oscillation.
The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on the surface of planet X such that the acceleration of the planet X is n times greater than the Earth is
A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according to as y = kt2, where y is the vertical distance covered and k = 1 ms−2, its time period becomes T2. Then, T `"T"_1^2/"T"_2^2` is (g = 10 ms−2)
Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is
T = `2π sqrt("R"/"g")`
Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth's surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is ______.
(consider the radius of earth RE = 6400 km and g on earth 10 m/s2)