SHM expression
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I. y=sin ωt−cos ωt
II. y=sin3 ωt
III. y=5 cos (3π4−3ωt)
IV. y=1+ω2t2+ωt
- Only (IV) does not represent S.H.M.
- (I) and (III)
- (I) and (II)
- Only (I)
- The motion is SHM.
- The motion is SHM with amplitude a+b.
- The motion is SHM with amplitude a2+b2.
- The motion is SHM with amplitude √a2+b2.
- 0.2 s
- 0.1 s
- 0.15 s
- 0.132 s
- 218 s
- 214 s
- 212 s
- 2110 s
- A2, πω, 2ω√(A−x)x
- 3A2, πω, ω√(A−x)x
- 3A2, πω, 3ω√(A−x)x
- A2, 2πω, 2ω√(A−x)x
- the amplitudes on both SHM should be equal and they should have a phase difference of π2
- the amplitudes should be in the ratio 1:2 and the phase difference should be zero
- the amplitudes should be in the ratio 1:2 and the phase difference should be π2
- the amplitudes should be equal and the phase difference should be zero
- π22 s
- 2π23 s
- 6 s
- π3 s
- a−b
- 2a−b3
- 2a23a−b
- None of these
- π3
- π2
- 2π3
- 4π5
- a2T2+4π2v2
- aT/x
- aT+2πv
- aT/v
- π3
- π8
- π6
- π4
The co-ordinates of a moving particle are and , where and are constants. the velocity of a particle at any instant is
Which of the following equations can be that of an SHM? Here, x represents displacement from mean position.
t represents time
k is a positive constant
v represents velocity
md2xdt2−kx=0
md2xdt2+kx=0
mdvdt+kx=0
K∫t0vdt=−md2xdt2 .
- a2T2+4π2v2
- aT/x
- aT+2πv
- aT/v
- √3π50^j m/s
- −√3π50^j m/s
- √3π50^i m/s
- −√3π50^i m/s
- π2
- π
- 3π2
- 0
A particle executing simple harmonic motion along y-axis has its motion described by the equation. y=Asin(ωt)+B. The amplitude of the simple harmonic motion is
A
B
A+B
√A+B
- x=a sin (ωt+ϕ)
- x=a cos (ωt+δ)
- x=a sin ωt+b cos ωt
- x=a sin ωt cos ωt
- 56 s
- 5π6 s
- 1.2 s
- 53 s
- 0.01
- 0.05
- 0.5
- 0.25
- The motion is SHM.
- The motion is SHM with amplitude a+b.
- The motion is SHM with amplitude a2+b2.
- The motion is SHM with amplitude √a2+b2.
[Take g=10 m/s2)]
- 2 rad/s
- 4 rad/s
- 6 rad/s
- 8 rad/s
A test tube of cross-sectional area a has some lead shots in it. The total mass is m. It floats upright in a liquid of density d. When pushed down a little and released, it oscillates up and down with a period T. Choose the correct relationship from the following.
T=2π√magd
T=2π√mgad
T=2π√mdag
T=2π√magd
- the amplitude of oscillation of 5 kg block is 0.18 m
- the first maximum compression occurs after start at 3π56 s
- the maximum extension of the spring is 0.50 m
- time period of oscillation is π14 s
- 5 cm
- 4 cm
- 2 cm
- 10 cm
- 5 Hz
- 6 Hz
- 7 Hz
- 8 Hz
- Function f(t) = sin kt + coskt is simple harmonic having a period 2πk
- Function f(t) = sin πt+2 cos 2π+3 sin 3 πt is periodic but not simple harmonic having a period of 2s
- Function f(t) = cos kt+2 sin2kt is simple harmonic having a period 2πk.
- Function f(t) = e−kt is not periodic.
- √3Aω2
- √3Aω4
- 3Aωπ
- 3Aωπ(2−√3)