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The term oscillation refers to the continuous to and fro movement of an object between two positions or states. A particle is said to have SHM i.e. simple harmonic motion if it moves to and fro about a fixed point such that its acceleration at any instant is directly proportional to displacement but the direction of acceleration is opposite to the direction of displacement. Example: motion of spring on a frictionless surface when its one end is fixed and other is in motion.
Read Also: Periodic Motion
Ques- The path length of oscillating simple pendulum of length 1 m is 16 cm its max velocity is ( g = π2 m/s2 )
- 2 π cm/s
- 4 π cm/s
- 8 π cm/s
- 16 π cm/s
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Ans: (C)
Explanation:
(mgl – mgl cosθ) = ½ mv2
mgl ( 1 – cosθ) = ½ mv2
m is cancelled from both sides
2gl ( 1 – cosθ) = v2
2gl sin2θ = v2 --------------------------------------- 1 – cosθ = sin2θ
v = √ (2 gl sin2θ)
Multiply and divide by 2
v = √ (4 g l sin2θ)/2
v = 2√ (g l sin2θ)/2
v = 2√ (g l sin2θ/4)
v = 2√ (g l) × sinθ/2
v = 2√ (g l) × θ/2 ------------------ sinθ/2 = θ/2
v = θ√ (g l)
We need to find θ

Ques- The distance covered by particle undergoing SHM in one time period is ------------- (Amplitude A)
- 4 A
- Zero
- A
- 2 A
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Ans: (A)
Explanation:
To understand the above problem, observed the figure carefully, in every half movement the particle covers distance A from its original position to right then come again its original position i.e. A +A = 2 A then pendulum goes left and covers distance A then come back its original position and covers distance A i.e. A+A =2A, it means total distance is 4 A

Ques– The dimension of k in equation F = kx is ----------------
- [ L0 M1 T-2 ]
- [ L2 M1 T-1 ]
- [ L2 M1 T-2 ]
- [ L0 M1 T-1 ]
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Ans: (A)
Explanation:
F = Force - [ L1 M1 T-2 ]
x – Distance – L1
F / x = k
k = L0 M1 T-2
Ques- A particle performing SHM of periodic time ‘T’. The time taken by the particle in moving from means point to half the maximum displacement is ------------ ( Sin 30 = 0.5 )
- T / 2
- T / 4
- T / 8
- T / 12
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Ans: (D)
Explanation-
x = A sin ωt (from mean point)
x= A
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x = A/2
A/2 = A sin ωt
½ = sin ωt
Sin-1 (1/2) = ωt
π/ 6 = 2π/T × t
t = T/12
Ques- Calculate the spring constant if the spring with 2N force is stretched by 40 cm
- -5 N/m
- 0 N/m
- 8 N/m
- -5 m/N
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Ans: (A)
Explanation-
F = -k x
F / x = -k
2N/0.4 m = -k
k = -5 N/ m
Ques- A particle is moving in a circle with uniform speed its motion is --------------
- Periodic and simple harmonic
- Periodic but not simple harmonic
- Non periodic
- None of the above
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Ans: (B)
Explanation-
If the speed of particle is uniform i.e. periodic in motion. But due to the retardation it will not be simple harmonic.
Ques- A small body of mass 0.10 kg is performing SHM of amplitude 1.0 m and period 0.20 S the maximum force acting on it is ...................
- 98.596 N
- 985.96 N
- 100.2 N
- 76.23 N
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Ans: (A)
Explanation-
Fmax = m amax ----------------- (1)
amax = Aω2
amax = 1× (2π / T)2
amax = 1× (4π2 / T2)
amax = (4π2 / 0.22) ---------------------------- π = 3.14
amax = 985.9 -------------------- put in (1)
Fmax = 0.1 × 985.9
Fmax = 98.59 N
Ques- The maximum velocity of particle in SHM is 0.16 m/s and maximum acceleration is 0.64 m/s2 . The amplitude is ------------------------
- 4×10-2 m
- 4×10-1 m
- 4×10 m
- 4×100 m
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Ans: (A)
Explanation –
Vmax = A ω --------------------------- (1)
amax = A ω2 ---------------------------- (2)
Divide (2) by (1)
amax/ Vmax = Aω2/ Aω
ω = 0.64 / 0.16
ω = 4 ................... put in ( 1 )
0.16 = A ×4
A = 0.04
A = 4×10-2 m
Ques- In a second pendulum mass of bob is 30 g if it is replaced by 90 g mass then its time period will be --------------------
- 1s
- 2s
- 4s
- 3s
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Ans: (B)
Explanation –
T = 2 π √ ( l / g)
The time period of second pendulum is always 2 sec
In this case you have to remember this condition.
Ques- The maximum velocity of a particle performing linear SHM is 0.16 m/s if its maximum acceleration is 0.64 m/s2 . Calculate its period
- 2 s
- 1.7 s
- 1.5 s
- 1 s
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Ans: (C)
Explanation –
T = 2 π / ω
Vmax = A ω --------------------------- (1)
amax = A ω2 ---------------------------- (2)
Divide (2) by (1)
amax/ Vmax = Aω2/ Aω
ω = 0.64 / 0.16
ω = 4
T = 2 π / 4
T = 1.5 s
Ques- Energy of a simple harmonic motion depends upon -----------
- 1 / ω2
- ω
- A2
- 1 / A2
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Ans: (C)
Explanation –
- E. = ½ K A2
- E. = ½ m ω2 A2
From the above equation we can conclude that the energy of a simple harmonic motion depends upon the square of amplitude.
Ques- The SHM of a particle is given by the equation x = 3 sin (ωt + ωt) the amplitude is -------------
- 1
- 2
- 3
- 4
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Ans: (C)
Explanation –
To find the amplitude we need to focus on following equation
Displacement (x) = A sin (ωt + ωt)
A is amplitude
A = 3
Ques- If the velocity of a body is half the maximum velocity. Then what is the distance from the mean position?
- 2 A
- √3 / 2 A
- A
- A / 2
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Ans: (B)
Explanation –
V = ω √(A2 – x2 )
V/2 = ω √(A2 – x2 )
Aω/2 = ω √(A2 – x2 )
A/2 = √(A2 – x2 )
Squaring both sides
A2/ 4 = (A2 – x2)
A2/ 4 - A2 = – x2
-3/4 A2 = – x2
Negative sign will be canceled
√3 / 2 A = x






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