Numerical Problems on S.H.M.

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Example – 1:

  • A particle performs a linear S.H.M along a path 10 cm long. The particle starts from a distance of 1 cm from the mean position towards the positive extremity. Find the epoch and the phase of motion when the displacement is 2.5 cm.
  • Solution:
  • Given:  Path length = 10 cm, amplitude = path length /2 = 10/2 = 5 cm, Initial displacement = x0 = 1 cm, Displacement = x = 2.5 cm.
  • To Find: Epoch = α =? and phase of S.H.M. = (ωt + α) = ?

Epoch = α = sin-1(xo/a) = sin-1(1/5) = sin-1(0.2) = 11°32’

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  2.5 = 5 sin (ωt + α)



∴  sin (ωt + α) = 2.5/5 = 1/2

∴  (ωt + α) = sin-1(1/2) = π/6

Ans: Initial phase is 11°32’ and phase of S.H.M. is π/6 or 30o.

Example – 2:

  • A particle performing S.H.M. has a period of 6 s and amplitude of 8 cm. The particle starts from the mean position and moves towards the positive extremity. Find its displacement, velocity, and acceleration 0.5 s after the start.
  • Solution:
  • Given:  Period = T = 6 s, amplitude = a = 8cm, time elapsed = t = 0.5 s, particle starts from mean position, α = 0.
  • To Find: Displacement = x = ?, Velocity = v =?, acceleration = f = ?

Angular velocity = ω = 2π/T = 2π/6 = π/3 rad/s



Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  x = 8 sin ( π/3 x 0.5 + 0)

∴  x = 8 sin ( π/6) = 8 x 1/2 = 4 cm

The magnitude of the velocity of a particle performing S.H.M. is given by



The magnitude of the acceleration of a particle performing S.H.M. is given by

f = ω2x = (π/3)2 x 4 =  (3.142/3)2 x 4 = 4.38 cm/s2.

Ans: Displacement = 4 cm; velocity = 7.26 cm/; acceleration = 4.38 cm/s2

Example – 3:

  • A particle performs S.H.M. with a period of 12 s. If its velocity is 6 cm/s two seconds after crossing the mean position, what is the amplitude of its motion?
  • Solution:
  • Given: Period = T = 12 s, v = 6 cm/s, time elapsed = t = 2 s, particle passes through mean position, α = 0.
  • To Find: amplitude = a =?

Angular velocity = ω = 2π/T = 2π/12 = π/6 rad/s



Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  x = a sin ( π/6 x 2 + 0)

∴  x = a sin ( π/3) = a √3/2   cm



Ans: The amplitude of motion is 22.92 cm

Example – 4:

  • A sewing machine needle moves in a path 4 cm long and the frequency of its oscillations is 10 Hz. What is its displacement and acceleration 1/120 s after crossing the centre of its path?
  • Solution:
  • Given: Path length = 4 cm, amplitude = path length/2 = 4/2 = 2 cm, Frequency of oscillation = n = 10 Hz, Time elapsed = t = 1/120 s,  particle passes through mean position, α = 0.
  • To Find: Displacement = x =? acceleration = f =?

Agular velocity = ω = 2πn = 2π x 10 = 20π rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)



∴  x = 2 sin ( 20π x 1/120 + 0)

∴  x = 2 sin ( π/6) = 2 x 1/2 = 1 cm

The magnitude of the acceleration of a particle performing S.H.M. is given by

f = ω2x = (20π)2 x 1 =  (20 x 3.142)2 = 3944 cm/s2.

Ans: Displacement =1 cm and acceleration = 3944 cm/s2

Example – 5:

  • A particle in S.H.M. has a velocity of 10 cm/s when it crosses the mean position. If the amplitude of its oscillations is 2 cm, find the velocity. When it is midway between the mean and extreme positions.
  • Solution: 
  • Given: Velocity at mean position = vmax = 10 cm/s, amplitude = a = 2 cm, Displacement midway between the mean and extreme positions, hence x = a/2 = 2/2 = 1 cm.
  • To Find: Velocity = v =?

We have vmax = ωa



∴  10 = ω x 2

∴  ω = 10/2 = 5 rad/s

Ans: velocity at midway between the mean and extreme positions is 8.66 cm/s



Example – 6:

  • Show that the velocity of a particle performing S.H.M. is half the maximum velocity at a displacement of √3/2 times its amplitude.
  • Solution:
  • Given: Displacement x = a√3/2
  • To Show: v = 1/2 vmax.

Example – 7:

  • The shortest distance traveled by a particle performing S.H.M. from its mean position in 2 seconds is equal to √3/2  of its amplitude. Find its period.
  • Solution:
  • Given: Time elapsed = t = 2s, displacement = x = a √3/2, particle passes through mean position, α = 0.
  • To Find: Period = T = ?

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  a √3/2  = a sin (ωt + 0)

∴  √3/2  = sin ωt

∴  ωt = sin-1(√3/2 ) = π/3



∴  (2π/T)t = π/3

∴  (2π/T)x 2 = π/3

∴  T = 2 x 2 x 3 = 12 s

Ans: Period is 12 s

Example – 8:

  • A particle performs a linear S.H.M. Its velocity is 3 cm/s when it is at 4 cm from the mean position and 4 cm/s when it is at 3 cm from the mean position. Find the amplitude and the period of S.H.M.
  • Solution:
  • Given: v1 = 3 cm/s at x1 = 4cm and v2 = 4 cm/s at x2 = 3cm
  • To Find: Amplitude = a =? Period = T=?

∴  16a2 -256 = 9a2 -81

∴  16a2 – 9a2  = 256  – 81

∴  7a2  = 175

∴  a2  = 25

∴  a = 5

Now T = 2π/ω = 2 x 3.14 /1 = 6. 28 s

Ans: Amplitude = 5 cm and period = 6.28 s

Example – 9:

  • The velocities of a particle performing linear S.H.M. are 0.13 m/s and 0.12 m/s when it is at 0.12 m and 0.13 m respectively from the mean position. Find its period and amplitude.
  • Solution:
  • Given: v1 = 0.13 m/s = 13 cm/s at x1 = 0.12 m = 12 cm and v2 = 0.12 m/s = 12 cm/s at x2 = 0.13 m = 13 cm
  • To Find: Amplitude = a =? Period = T=?

∴  144a2 – 144 x 144 = 169a2 – 169x 169

∴  169a2 – 144a2  = 169 x 169 – 144x 144

∴  25a2  = (169 + 144)(169 – 144)

∴  25a2  = (313)(25)

∴  a2 = 313

∴  a = √313 = 17.69 m

Now T = 2π/ω = 2 x 3.14 /1 = 6. 28 s

Ans:  Period = 6.28 s and amplitude = 17.69 cm

Example – 10:

  • A particle performing  S.H.M. has velocities of 8 cm/s and 6 cm/s at displacements of 3 cm and 4 cm respectively. Find its amplitude and frequency of oscillations. Calculate its maximum velocity. What is the phase of its motion when the displacement is 2.5 cm?
  • Solution:
  • Given: v1 = 8 cm/s at x1 =3 cm and v2 = 6 cm/s at x2 = 4 cm, displacement = x = 2.5 cm
  • To Find: Amplitude = a =? frequency = n = ?, phase = (ωt + α) =?,

∴  9a2 – 81 = 16a2 – 256

∴  16a2 – 9a2  = 256 – 81

∴  7a2  = 175

∴  a2  = 25

∴  a = 5 cm

Now ω = 2 π n

∴ n = ω/2π = 2/( 2 x 3.142) = 0.3183 Hz

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  2.5 = 5 sin (ωt + α)

∴  sin (ωt + α) = 2.5/5 = 1/2

∴  (ωt + α) = sin-1(1/2) = π/6

Ans: Amplitude is 5 cm, frequency = 0.3183 Hz, Phase = π/6 or 30°

Example – 11:

  • The periodic time of a body executing S.H.M. is 2 s. After how much time interval from t =0 will its displacement be half the amplitude?
  • Solution:
  • Given: Time period = T = 2 s, Displacement x = 1/2 a, particle passes through mean position, α = 0.
  • To Find: Time elapsed = t =?

Angular velocity = ω = 2π/T = 2π/2 = π rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  1/2 a = a sin (πt + 0)

∴  1/2 =sin (πt)

∴  πt = sin-1(1/2) = π/6

∴  t = 1/6 s

Ans: After 1/6 s the displacement will be half the amplitude

Example – 12:

  • A particle executes S.H.M.  of amplitude 25 cm and time period 1/3 seconds. What is the minimum time required for a particle to move between two points located at a distance of 12.5 cm on either side the mean position?
  • Solution:
  • Given: amplitude = a = 25 cm, x = 12.5 on either side, Period = T = 3 s, particle passes through mean position, α = 0.
  • To Find: Time required = 2t = ?

Angular velocity = ω = 2π/T = 2π/3  rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  12.5 = 25 sin ((2π/3)t + 0)

∴  1/2 =sin ((2π/3)t)

∴  (2π/3)t = sin-1(1/2) = π/6

∴  t = 1/4 s

Now the points are on either side of the mean position

Hence time taken to move between these two points = 2t = 2 x 1/4 = 0.5 s

Ans: the minimum time required is 0.5 s

Example – 13:

  • A particle executes S.H.M. of period 12 s and of amplitude 8 cm. What time will it take to travel 4 cm from the extreme position ?
  • Solution:
  • Given: Period = T = 12 s, amplitude = a = 8cm, distance from extreme position = 4 cm, displacement = x = 8 cm – 4 cm = 4 cm, particle starts from extreme position, α = π/2.
  • To Find: time taken = t = ?, Velocity = v = ?

Angular velocity = ω = 2π/T = 2π/12  = π/6 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  4 = 8 sin ((π/6)t + π/2)

∴  1/2 =cos ((π/6)t)

∴  (π/6)t = cos-1(1/2) = π/3

∴  t = 2 s

Ans: Time taken = 2 cm

Example – 14:

  • A particle performs S.H.M. of period 4 s. If the amplitude of its oscillations is 4 cm, find the time it takes to describe 1 cm from the extreme position.
  • Solution:
  • Given: Period = T = 4 s, amplitude = a = 4cm, distance from extreme position = 1 cm, displacement = x = 4 cm – 1 cm = 3 cm, particle starts from extreme position, α = π/2.
  • To Find: time taken = t = ?

Angular velocity = ω = 2π/T = 2π/4  = π/2 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  3 = 4 sin ((π/2)t + π/2)

∴  3/4 = cos ((π/2)t)

∴  (π/2)t = cos-1(3/4) = 41.41° = 41.41 x 0.0175= 0.7247 rad

∴  t = 0.7247 x 2 /3.142 = 0.4613 s

Ans:  Time taken = 0.4613 s

Example – 15:

  • A particle performs S.H.M. of period 12 s along a path 16 cm long. If it is initially at the positive extremity, how much time will it take to cover a distance of 6 cm from that position?
  • Solution:
  • Given: Period = T = 12 s, path length = 16 cm, amplitude = a = 16/2 = 8cm, distance from extreme position = 6 cm, displacement = x = 8 cm – 6 cm =2 cm, particle starts from extreme position, α = π/2.
  • To Find: time taken w.r.t. extreme position = te = ?

Angular velocity = ω = 2π/T = 2π/12  = π/6 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  2 = 8 sin ((π/6)t + π/2)

∴  2/8 = cos ((π/6)t)

∴  (π/6)t = cos-1(1/4) = 75.52° =75.52 x 0.0175 = 1.322 rad

∴  t = 1.322 x 6 /3.142 = 2.52 s

Ans: Time taken = 2.52 s

Example – 16:

  • If a particle performing S.H. M. starts from the extreme position after an elapse of what fraction of the period will the velocity of the particle be half the maximum velocity?
  • Solution:
  • Given: v = 1/2 vmax.   particle starts from extreme position, α = π/2.
  • Fo Find: Fraction of time = t/T =?

∴  4a2 – 4x2 = a2

∴   4x2 = 3a2

∴   2x = a√3

∴  x = a√3/2

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  a√3/2 = 1 sin ((2π/T)t + π/2)

∴  3/2 = cos ((2π/T)t)

∴  (2π/T)t = cos-1(3/2) = π/6

∴  t /T = 1/12 s

Ans: fraction of the period is 1/12 s

Example – 17:

  • a particle executing S.H.M. has a period of 6 s and its maximum velocity during oscillations is 6.28 cm/s. Find the time taken by it to describe a distance of 3 cm from its equilibrium position.
  • Solution:
  • Given: Period = T = 6 s, Vmax = 6.28 cm/s, x = 3 cm, particle passes through mean position, α = 0.
  • To Find: Time taken = t =?

Angular velocity = ω = 2π/T = 2π/6  = π/3 rad/s

vmax = ωa

∴  a = vmax/ω  = 6.28 /(π/3) = 6 cm

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  3 = 6 sin ((π/3)t + 0)

∴  3/6 = sin ((π/3)t)

∴  (π/3)t = sin-1(1/2) = π/6

∴  t = 1/2 s = 0.5 s

Ans: Time taken = 0.5 s

Example – 18:

  • A particle performs S.H.M. of amplitude 10 cm. Its maximum velocity during oscillations is 100 cm/s. What is its displacement when the velocity is 60 cm/s?
  • Solution:
  • Given: amplitude = 10 cm, Vmax = 100 cm/s, v = 60 cm/s
  • To Find: displacement = x =?

vmax = ωa

∴  ω = vmax/a  = 100/10 = 10 rad/s

Ans: Displacement = 8 cm

Example – 19:

  • The maximum velocity of a particle performing S.H.M. is 6.28 cm/s. If the length of its path is 8 cm, calculate its period.
  • Solution:
  • Given: path length = 8 cm, amplitude = 8/2 = 4 cm, Vmax = 6.28 cm/s,
  • To Find: Period = T =?

vmax = ωa

∴  ω = vmax/a  = 6.28/4 = 1.57 rad/s

T = 2π /ω = (2 x 3.14)/ 1.57 = 4 s

Ans: Period = 4 s

Example – 20:

  • A particle performs S.H.M. of amplitude 3 cm. If its acceleration in the extreme position is 27 cm/s2, find the period.
  • Solution:
  • Given: Amplitude = a = 3 cm, acceleration at extreme position = f = 27 cm/s2,
  • To Find: Period = T =?

At extreme position acceleration is maximum, fmax = 27 cm/s2

fmax = ω2a

∴  ω2 = fmax/a  = 27/3 = 9

∴  ω  = 3 rad/s

T = 2π /ω = (2 x 3.14)/ 3 = 2.09 s

Ans: Period = 2.09 s

Example – 21:

  • A particle executing S.H.M. has a maximum velocity of 0.16 cm/s and a maximum acceleration of 0.64 m/s2. Calculate its amplitude and the period of oscillations.
  • Solution:
  • Given: vmax = 0.16 cm/s, f max = 0.64  m/s2.
  • To Find: Amplitude = a =? and Period = T = ?

fmax = ω2a ………. (1)

vmax = ωa  ………. (2)

Dividing equation (1) by (2)

fmax /vmax  = ω

∴  ω = 0.64/0.16 = 4 rad/s

Substituting in equation (2)

0.16 = 4 x a

∴ a = 0.04 cm

T = 2π /ω = (2 x 3.14)/ 4 = 1.57 s

Ans: amplitude = 0.04 cm and period = 1.57 s

Example – 22:

  • A particle performing S.H.M. along a straight line has a velocity of 4π cm/s when its displacement is √12 cm. If the maximum acceleration it can attain is 16πcm/s2, find the amplitude and the period of its oscillations.
  • Solution:
  • Given: vmax = 4π cm/s, f max = 16πm/s, Displacement = √12 cm
  • To Find: Amplitude = a =? and Period = T = ?

fmax = ω2a

∴  16π = ω2a  ………. (2)

From equations (1) and (2) we have

ω2(a2 – 12) =  ω2a

∴  (a2 – 12) =  a

∴  a2 – 12 – a = 0

∴ (a  – 4)(a + 3) = 0

∴  a = 4 cm or a = – 3 cm

Amplitude is maximum displacement hence a = 3 cm <  √12 cm is not possible.

∴  a = 4 cm

substituting in equation (2)

16π = ω2(4)

 ∴   ω= 4π2

 ∴   ω = 2π rad/s

∴   T = 2π /ω = 2π /2π =  1 s

Ans: amplitude = 4 cm and period = 1 s

Example – 23:

  • A particle of mass of 10 g performs S.H.M. of period 5 s and has an amplitude of 8 cm. Find its velocity when it is at a distance of 6 cm from the equilibrium position. Find also the maximum velocity and maximum force acting on it.
  • Solution:
  • Given: mass = m = 10 g, Period = T = 5 s, amplitude = a = 8 cm, displacement = x = 6 cm, particle passes through mean position, α = 0.
  • To Find: velocity = v = ?, vmax = ?, Fmax = ?

Angular velocity = ω = 2π/T = 2π/5  rad/s

Problems on S.H.M.

∴   vmax = ωa  = 2π/5 x 8 = 10.05 cm/s

fmax = ω2a = ( 2π/5)2 x 8  = 12.63 cm/s2

Fmax = m. fmax = 10 x 12.63 = 126.3 dyne

∴   Fmax =  126.3 x 10-5 N = 1.263 x 10-3 N

Ans: velocity =6 .65 cm/s;  maximum velocity =10.05 cm/s;  maximum force = 1.263 x 10-3 N

Example – 24:

  • A block is on a piston which is moving vertically up and down with S.H.M. of period one second. At what amplitude of motion will the block and piston separate? At which point in the path of motion will the separation take place?
  • Solution:
  • Given: Period = T = 1s
  • To Find: amplitude = a = ?

Angular velocity = ω = 2π/T = 2π/1 =  2π rad/s

At the topmost point, the block and piston will separate.

At topmost point acceleration is maximum. Hence force is maximum

Maximum force on the block = weight of the block

m. fmax =  mg

∴   fmax =  g

∴   ω2a =  g

∴   a =  g / ω2  = 980/ (2 x 3.142)2 = 24.82 cm

Ans: At amplitude = 24.82 cm blockw will separate at topmost point of the path

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