Advertisements
Advertisements
Question
A small block oscillates back and forth on a smooth concave surface of radius R in Figure. Find the time period of small oscillation.
Advertisements
Solution
It is given that R is the radius of the concave surface.
Let N be the normal reaction force.
Driving force, F = mg sin θ
Comparing the expression for driving force with the expression, F = ma, we get:
Acceleration, a = g sin θ
Since the value of θ is very small,
∴ sin θ → θ
∴ Acceleration, a = gθ
Let x be the displacement of the body from mean position.
\[\therefore \theta = \frac{x}{R}\]
\[ \Rightarrow a = g\theta = g\left( \frac{x}{R} \right)\]
\[ \Rightarrow \left( \frac{a}{x} \right) = \left( \frac{g}{R} \right)\]
\[\Rightarrow a = x\frac{g}{R}\]
As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.
Time period \[\left( T \right)\] is given by,
\[T = 2\pi\sqrt{\frac{\text { displacement }}{\text { Acceleration }}}\]
\[= 2\pi\sqrt{\frac{x}{gx/R}} = 2\pi\sqrt{\frac{R}{g}}\]
APPEARS IN
RELATED QUESTIONS
Show variation of displacement, velocity, and acceleration with phase for a particle performing linear S.H.M. graphically, when it starts from the extreme position.
Can a pendulum clock be used in an earth-satellite?
The displacement of a particle is given by \[\overrightarrow{r} = A\left( \overrightarrow{i} \cos\omega t + \overrightarrow{j} \sin\omega t \right) .\] The motion of the particle is
The average energy in one time period in simple harmonic motion is
A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives correct time at the equator. If the clock is taken to the poles it will
A pendulum clock keeping correct time is taken to high altitudes,
A particle moves in a circular path with a continuously increasing speed. Its motion is
For a particle executing simple harmonic motion, the acceleration is proportional to
A pendulum clock giving correct time at a place where g = 9.800 m/s2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.
A small block oscillates back and forth on a smooth concave surface of radius R ib Figure . Find the time period of small oscillation.
A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.
A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the road.
Three simple harmonic motions of equal amplitude A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion.
A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45° with the X-axis. The two motions are given by x = x0 sin ωt and s = s0 sin ωt. Find the amplitude of the resultant motion.
A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3 s. The time period is ____________.
Define the time period of simple harmonic motion.
Write short notes on two springs connected in parallel.
Consider two simple harmonic motion along the x and y-axis having the same frequencies but different amplitudes as x = A sin (ωt + φ) (along x-axis) and y = B sin ωt (along y-axis). Then show that
`"x"^2/"A"^2 + "y"^2/"B"^2 - (2"xy")/"AB" cos φ = sin^2 φ`
and also discuss the special cases when
- φ = 0
- φ = π
- φ = `π/2`
- φ = `π/2` and A = B
- φ = `π/4`
Note: when a particle is subjected to two simple harmonic motions at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.
Displacement vs. time curve for a particle executing S.H.M. is shown in figure. Choose the correct statements.
- Phase of the oscillator is same at t = 0 s and t = 2s.
- Phase of the oscillator is same at t = 2 s and t = 6s.
- Phase of the oscillator is same at t = 1 s and t = 7s.
- Phase of the oscillator is same at t = 1 s and t = 5s.
If x = `5 sin (pi t + pi/3) m` represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are ______.
