Phase and the General Equation
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A simple harmonic motion is represented by . The angular frequency of oscillation is given by?
- 2 ms
- 0.3 ms
- 4 ms
- 3 ms
- T4
- T8
- T12
- T2
- T8
- 3T8
- T6
- 4T3
- π60 s
- π120 s
- π30 s
- 5π3 s
- Zero
- π/4
- π/2
- π
- Zero
- π2
- π
- 2π
A particle executing simple harmonic motion has angular frequency 6.28 s−1 and amplitude 10 cm. Find (x) the speed when the displacement is 6 cm, from the mean position, (y) the speed at t=16 s assuming that the motion starts from rest at t=0.
(i) 25.1 cms−1
(ii) 54.4 cms−1
(iii) 50.2 cms−1
(iv) 27.2 cms−1
x - (ii); y - (iv)
x - (iii); y - (iv)
x - (i), y - (iii)
x - (iii), y - (ii)
- π
- cos−1(23)
- π3
- cos−1(79)
78.8m/s2
- 100m/s2
- 50m/s2
- 40m/s2
- 124 s
- 116 s
- −116 s
- 316 s
- T2
- T4
- T8
- T12
- t=0
- t=π4ω
- t=π2ω
- t=πω
A particle is undergoing circular motion with angular frequency ω as shown
If the shadow of the particle on the wall is performing SHM such that its acceleration's maximum value is 'a', find the acceleration of the particle and the radius of the circle.
a, 92ω2
2a, aω2
a, aω2
Data is sufficient
- T1 <T2
- T1>T2
- T1=T2
- T1=2T2
The equation of a particle executing SHM is x=(5m)sin[(π s−1)t+π3]. What is its amplitude, time period, maximum speed and velocity. At t = 1s?
5m, 2s, 5ms−1, −5π2ms−1
5m, 2s, 5π ms−1, −5π2ms−1
5m, πs, 5 ms−1, −5π2ms−1
5m, 2s, 5π ms−1, +5π2ms−1
- 0
- π
- π2
- 2π
- T
- T4
- T8
- T16
- x=5 cos ωt
- x=−5 cos ωt
- x=5 sin ωt
- x=10 cos ωt
The phase difference between two SHM
y1=10sin(10π t+π3) and y2=12sin(8π t+π4) at t=0.5s is:
11π12
13π12
π
17π12
- x=Asin(ωt+5π6)
- x=Asin(ωt+π6)
- x=Asin(ωt−5π6)
- x=Asin(ωt−π6)
- cos−1(79)
- cos−1(59)
- cos−1(49)
- cos−1(19)
A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, find the phase difference between the individual motions.
0
(π3)
(2π3)
(π2)
Corresponding to point A on acceleration time (a-t) curve of a particle executing SHM, there is a point 1 or 2 or 3 or 4 on velocity time (v-t) curve of the motion. The starting value of time on x-axis is different for two curves. Find the point:
1
2
3
4
- 3:1
- 1:3
- 1:9
- 9:1
detached from it ?
- g4π2v2
- mg4π2v2
- g2π2v2
- gm2π2v2
A spring block system is doing SHM with Amplitude A and angular frequency ω.
At t=T, the particle is at extreme position. Write displacement versus time equation.
x=Asin ω t
x=A+Asin ω t
x=Acos ω t
x=Asin(ω t+Π2)
- ±0.58 m/s, 0.45 m/s2
- ±0.8 m/s, 1.8 m/s2
- ±0.58 m/s, −0.45 m/s2
- ±0.8 m/s, 0.45 m/s2
- T
- T4
- T8
- T16