Simple Harmonic Motion (SHM) Numericals Class 12

Numerical Problems:

Type – 1 (Simple Pendulum) 

(i) Time period, T = 2$\pi $ $\sqrt{\frac{l}{g}}$

 

Where l is the length of pendulum.

(ii) For second pendulum, T = 2 sec, 

(iii) At mean position:

Displacement, y = 0 (minimum)

Velocity = Maximum

Acceleration, a = Minimum

(iv) At extreme positions:

Displacement, y = r (maximum)

Velocity = 0 (minimum)

Acceleration, a = Maximum

In S.H.M.

(v) Velocity, v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

where, $\omega $ is angular velocity when uniform circular motion is taken as reference of S.H.M.

r is amplitude

y is displacement

(v) Acceleration, a = -$\omega $² y     

(vi) Time period, T = $\frac{2\pi }{\omega }$

Q.1. A simple pendulum 4m long swings with an amplitude of 0.2m. Compute the velocity of the pendulum at its lowest point and its acceleration at extreme ends.

Solution: 

Here,

Length, l = 4m 

Amplitude, r = 0.2m

(i)  At lowest point (mean position), velocity, v = ? 

(ii) At extreme ends, acceleration, a = ? 

We have, 

For simple pendulum, Time period 

T = 2$\pi $ $\sqrt{\frac{l}{g}}$ = 2$\pi $ $\sqrt{\frac{4}{10}}$ = 3.974 s

(i)  Velocity, v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

But at lowest point (mean position), y = 0

Or,   v = $\frac{2\pi }{T}$ × r    since, $\omega $ = $\frac{2\pi }{T}$

Or,   v = $\frac{2\pi }{3.974}$× 0.2 =   0.316 ms^–1 

(ii) Acceleration, a = $\omega $² y  [taking magnitude only]

 At extreme ends, y = r

Or,   a = $\omega $² r  

Or,   a = ${{\left( \frac{2\pi }{T} \right)}^{2}}$× r  

$\therefore $  a = ${{\left( \frac{2\pi }{3.974} \right)}^{2}}$× 0.2 = 0.5 ms^–2 

Q.2. A simple pendulum has a period of 4.2 second, when the pendulum is shortened by 1m the period is 3.7 second. From these measurements, calculate the acceleration of free fall and the original length of the pendulum.

Solution:

Here, 

Case -1: 

Time period, T₁ = 4.2 sec

Length, l₁ = l m (say)

Case -2: 

Time period, T₂ = 3.7 sec

Length, l₂ = ( l – 1) m

Acceleration of free fall, g = ?

Original length, l = ?

We have, 

Time period, T = 2$\pi $$\sqrt{\frac{l}{g}}$

For 1^st case, T₁ = 2$\pi $$\sqrt{\frac{{{l}_{1}}}{g}}$

Or, 4.2 = 2$\pi $ $\sqrt{\frac{l}{g}}$……. (1)

For 2^nd case, T₂ = 2$\pi $ $\sqrt{\frac{{{l}_{2}}}{g}}$

Or, 3.7 = 2$\pi $ $\sqrt{\frac{l-1}{g}}$……. (2)

Dividing (1) by (2), we get

 $\frac{4.2}{3.7}$= $\sqrt{\frac{l}{l-1}}$

Squaring both sides,

Or,  $\frac{{{4.2}^{2}}}{{{3.7}^{2}}}$= $\frac{l}{l-1}$

Or, 1.29l – 1.29 = l

Or, l(1.29 – 1) = 1.29

Or, l = $\frac{1.29}{1.29-1}$

$\therefore $ Original length, l = 4.45m

From (1)

4.2 = 2$\pi $ $\sqrt{\frac{4.45}{g}}$

Or, g = ${{\left( \frac{2\pi }{4.2} \right)}^{2}}$× 4.45 

$\therefore $ Acceleration of free fall, g = 9.96 ms^–2

Q.3. A simple pendulum 5m long swings with an amplitude of 25cm. Compute the velocity of the pendulum at its lowest point and the acceleration at the end of its path.

Solution: 

Here,

 

Length, l = 5m, 

Amplitude, r = 25cm = 0.25m

(i) At lowest point (i.e. mean position), velocity, v = ?

(ii) At end points, acceleration, a = ? 

We have,

Time period, T = 2$\pi $ $\sqrt{\frac{l}{g}}$ 

Or, T = 2$\pi $$\sqrt{\frac{5}{10}}$ = 4.44 secs

Now,

(1) Velocity, v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

At mean position, y = 0

Or, v = $\frac{2\pi }{T}$ . r = $\frac{2\pi }{4.44}$ × 0.25 

$\therefore $ v = 0.354 m/s

(2) Acceleration, a = $\omega $² y        [Taking magnitude only]

At extreme positions, y = r

Or, a = ${{\left( \frac{2\pi }{T} \right)}^{2}}$× r 

Or, a =${{\left( \frac{2\pi }{4.44} \right)}^{2}}$× 0.25 

$\therefore $ a = 0.5 m/s²

Q.4. A second pendulum is taken to the moon. If the time period on the surface of moon is 4.90 seconds, what will be the acceleration due to gravity of the moon? 

Solution:

Acceleration due to gravity of the moon, g_m = ?

Case -1: On the earth,

Time period for second pendulum, T₁ = 2sec

T₁ = 2$\pi $ $\sqrt{\frac{l}{{{g}_{e}}}}$

Or, 2 = 2$\pi $ $\sqrt{\frac{l}{{{g}_{e}}}}$…….  (1)

Case -2: When pendulum is taken to the moon, 

Time period, T₂ = 4.9sec

T₂ = 2$\pi $ $\sqrt{\frac{l}{{{g}_{m}}}}$

Or, 4.9 = 2$\pi $ $\sqrt{\frac{l}{{{g}_{m}}}}$…….  (2)

Dividing equation (2) by (1) 

$\frac{4.9}{2}$= $\sqrt{\frac{{{g}_{e}}}{{{g}_{m}}}}$

Or, g_m = $\frac{{{2}^{2}}}{{{4.9}^{2}}}$× 9.8 

$\therefore $  g_m = 1.633 m/s²

Q.5. Calculate the period of oscillation of a simple pendulum of length 1.8m with a bob of mass 2.2kg. If the bob of this pendulum is pulled aside a horizontal distance of 20cm and released. What will be the value of (i) the K.E. and (ii) the velocity of the bob at the lowest point of the swing?

Solution: 

Here,

Time period of oscillation, T = ?

Length of pendulum, l = 1.8m, 

Mass of bob, m = 2.2kg, 

Amplitude, r = 0.2m

At lowest (Mean position) (i) K.E = ?   and   (ii) v = ?

We have,

Time period, T = 2$\pi $ $\sqrt{\frac{l}{g}}$ 

Or, T = 2$\pi $$\sqrt{\frac{1.8}{10}}$= 2.66 sec

(ii) v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

At mean position, y = 0

Or, Velocity, v = $\frac{2\pi }{T}$ . r = $\frac{2\pi }{2.66}$ × 0.2 = 0.47 m/s

(i) K.E = $\frac{1}{2}$ mv² = $\frac{1}{2}$×2.2×0.47² = 0.243 J

Type – 2: (Motion of spring-mass system)

Hooke’s law, the restoring force is directly proportional to displacement. i.e.

      F $\propto $ x   [Or, F $\propto $ y    Or,    F $\propto $ l] 

Or, F = -kx    Where k is called force constant or spring constant.

# Motion of Horizontal spring-mass system on frictionless surface.

Hooke’s law, 

      F $\propto $ x 

Or, F = -kx    

Time Period (T):

T = 2$\pi $ $\sqrt{\frac{m}{k}}$    

Also,

$\omega $ = $\sqrt{\frac{k}{m}}$

# Motion of a vertically loaded spring.

Fig: Motion of vertically loaded spring

Hooke’s law (In fig: b)

              F₁ $\propto $ l

Or,  F₁ = -kl = mg

Hooke’s law (In fig: c)

F₂  $\propto $ (l + y)

 F₂ = -k (l + y) = mg + ma 

Time Period (T):

T = 2$\pi $ $\sqrt{\frac{m}{k}}$        

  Also, $\omega $ = $\sqrt{\frac{k}{m}}$

Q.6. A body of mass 0.1 kg is undergoing simple harmonic motion of amplitude 1 m and period 0.2 second. If the oscillation is produced by a spring what will be the maximum value of the force and the force constant of the spring?

Here,

Mass, m = 0.1 kg

Amplitude, r = 1 m

Time period, T = 0.2 sec

Maximum force, F_max = ?

Force constant of spring, k = ?

(1) Maximum force, F_max = m a_max

Or, F_max = m×($\omega $²y)_max 

Or, F_max = m$\omega $²r [since, y_max = r]

Or, F_max = m×${{\left( \frac{2\pi }{T} \right)}^{2}}$×r 

Or, F_max = 0.1×${{\left( \frac{2\pi }{0.2} \right)}^{2}}$×1 = 98.7 N

Again,

(2) Time period, T = 2$\pi $ $\sqrt{\frac{m}{k}}$        

Or, k = ${{\left( \frac{2\pi }{T} \right)}^{2}}$×m

$\therefore $  k = ${{\left( \frac{2\pi }{0.2} \right)}^{2}}$×0.1 = 98.7 Nm^–1

Q.7. A body of mass 2 kg is suspended from a spring of negligible mass and is found to stretch the spring 0.1 m. What is its force constant and the time period?

Here,

Mass, m = 2 kg

Elongation, l = 0.1 m 

(1) Force constant of spring, k = ?

(2) Time period, T = ?

Here, Applied force = Restoring force 

Or, mg = kl (taking magnitude only)

Or, k = $\frac{mg}{l}$= $\frac{2\times 10}{0.1}$= 200 Nm^–1

Again,

(2) Time period, T = 2$\pi $ $\sqrt{\frac{m}{k}}$        

$\therefore $  T = 2$\pi $ $\sqrt{\frac{2}{200}}$ = 0.628 sec

Q.8. A glider with mass m = 2.00 kg sits on a frictionless air track, connected to a spring with force constant k = 5.00 N/m. You pull the glider, stretching the spring 0.100 m and then releases it with no initial velocity. The glider begins to move back towards its equilibrium position (x = 0). What is its velocity when x = 0.080m?

Here,

Mass, m = 2 kg

Force constant, k =5 Nm^–1

Amplitude, r = 0.1 m

Velocity, v = ? at displacement, x = 0.08 m 

Time period, T = 2$\pi $ $\sqrt{\frac{m}{k}}$        

Or, T = 2$\pi $ $\sqrt{\frac{2}{5}}$ = 3.974 sec

Now, Velocity, v = $\omega $$\sqrt{{{r}^{2}}-{{x}^{2}}}$

Or, v = $\frac{2\pi }{T}$ ×$\sqrt{{{r}^{2}}-{{x}^{2}}}$

Or, v = $\frac{2\pi }{3.974}$ × $\sqrt{{{0.1}^{2}}-{{0.08}^{2}}}$

$\therefore $  v = 0.095 ms^–1

Q.9. A small mass of 0.2 kg is attached to one end of helical spring and produces an extension of 15mm. The mass is now set into vertical oscillation of amplitude 10 mm. What is:

i. the period of oscillation?

ii. the maximum kinetic energy of the mass?

iii. the potential energy of the mass, when the mass is 5mm below the centre of oscillation? 

(g = 9.8ms^–2)

Here,

Mass, m = 0.2 kg

Elongation produced, l = 15 mm = 15×10^–3 m

Amplitude of oscillation, r = 10 mm = 10×10^–3 m

(i) Time period of oscillation, T = ?

(ii) Maximum kinetic energy, K.E._max = ?

(iii) Potential energy, P.E. = ? when displacement, y = 5mm

(g = 9.8ms^–2)

When small mass of 0.2 kg is attached to one end of helical then

Applied force = Restoring force 

Or, mg = kl (taking magnitude only)

Or, k = $\frac{mg}{l}$= $\frac{0.2\times 9.8}{15\times {{10}^{3}}}$= 130.67 Nm^–1

(i) Time period, T = 2$\pi $ $\sqrt{\frac{m}{k}}$        

Or, T = 2$\pi $ $\sqrt{\frac{0.2}{130.67}}$ = 0.246 sec

(ii) In S.H.M. 

Maximum K.E. = Total energy 

i.e. K.E._max = T.E. = $\frac{1}{2}$ m $\omega $² r² 

K.E._max = $\frac{1}{2}$ m×${{\left( \frac{2\pi }{T} \right)}^{2}}$× r²   

K.E._max = $\frac{1}{2}$ × 0.2 ×${{\left( \frac{2\pi }{0.246} \right)}^{2}}$× (10×10^–3)²   

K.E._max = 6.5×10^–3 J

(iii) Also, P.E. = $\frac{1}{2}$ m$\omega $²y²

P.E. = $\frac{1}{2}$ m×${{\left( \frac{2\pi }{T} \right)}^{2}}$×y²   

P.E. = $\frac{1}{2}$ ×.02×${{\left( \frac{2\pi }{0.246} \right)}^{2}}$× (5×10^–3)²   

$\therefore $  P.E. = 1.63 ×10^–3 J

Type – 3 (General concept of S.H.M.)

(1) Displacement equation, y = r sin $\omega $t

In terms of phase angle, displacement, y = r sin ($\omega $t+$\phi $)       

(i) At mean position, displacement, y = 0

(ii) At extreme positions, displacement, y = r

(2)  Amplitude (r): = y_max = r

(3) Velocity, v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

 (i) At mean position, v = $\omega $r (maximum velocity) 

(ii) At extreme positions, v = 0 (minimum velocity)

(4) Acceleration, a = $\omega $² y 

 (i) At mean positions, a = 0 (minimum)

(ii) At extreme positions, a = $\omega $² r (maximum)

(5) Time Period, T = $\frac{2\pi }{\omega }$                                

(6) Frequency, f = $\frac{1}{T}$

(7) Total energy of a simple harmonic oscillator, T.E. = $\frac{1}{2}$ m$\omega $² r²        

(8)  P.E. = $\frac{1}{2}$ m$\omega $²y² 

(9) K.E. = $\frac{1}{2}$ m$\omega $² (r² – y²)

(i) At mean position, P.E = 0 and K.E = T.E.

(ii) At extreme positions, K.E = 0 and P.E.= T.E  

Q.10. A body of mass 200 gm is executing simple harmonic motion with amplitude of 20 mm. The maximum force which acts upon it is 0.8N. Calculate its maximum velocity and its period of oscillation.

Here,

Mass, m = 200 g = 0.2 kg

Amplitude, r = 20 mm = 20×10^–3 m

Maximum force, F_max = 0.8 N

(i) Maximum velocity, v_max =?

(ii) Time period, T = ?

Given

F_max = 0.8 N

Or, m a_max = 0.8

Or, m $\omega $²r = 0.8

Or, m ${{\left( \frac{2\pi }{T} \right)}^{2}}$r = 0.8

(2) T = $\sqrt{\frac{mr}{0.8}}$× 2$\pi $

Or, T = $\sqrt{\frac{0.2\times 20\times {{10}^{3}}}{0.8}}$× 2$\pi $

Or, T = 0.444 sec

(1) v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

Since v is maximum at mean position where y = 0

Or, v_max = $\omega $$\sqrt{{{r}^{2}}-{{0}^{2}}}$

Or, v_max = $\omega $r

Or, v_max = $\frac{2\pi }{T}$ r

Or, v_max = $\frac{2\pi }{0.444}$ × 20×10^–3 

$\therefore $         v_max =  0.283 ms^–1

Alternative method,

v_max = $\omega $r = $\frac{2\pi }{T}$ r = $\frac{2\pi }{0.444}$ × 20×10^–3 

$\therefore $    v_max = 0.283 ms^–1

Q.11. A body is vibrating with simple harmonic motion of amplitude 15cm and frequency 4 Hz. Calculate the maximum value of acceleration and velocity.

Here,

Amplitude, r = 15 cm = 0.15 m

Frequency, f = 4 Hz

(1) Maximum acceleration, a_max =?

(2) Maximum velocity, v_max =?

We know, time period, T = $\frac{1}{f}$ = $\frac{1}{4}$ = 0.25 sec

(1) Acceleration, a = $\omega $²y

$\therefore $         a_max = ($\omega $²y)_max 

Or,        a_max = $\omega $²y_max 

Or,        a_max =  ${{\left( \frac{2\pi }{T} \right)}^{2}}$× r

Or,        a_max =  ${{\left( \frac{2\pi }{0.25} \right)}^{2}}$× 0.15 = 94.75 ms^–2 

Alternative method:

a_max = $\omega $²r =  ${{\left( \frac{2\pi }{T} \right)}^{2}}$× r =  ${{\left( \frac{2\pi }{0.25} \right)}^{2}}$× 0.15 = 94.75 ms^–2 

(2)       v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

Since v is maximum at mean position where y = 0

v_max = $\omega $$\sqrt{{{r}^{2}}-{{0}^{2}}}$

Or, v_max = $\omega $r

Or, v_max = $\frac{2\pi }{T}$ r

Or, v_max = $\frac{2\pi }{0.25}$ × 0.15 = 3.77 ms^–1

Alternative method,

v_max = $\omega $r = $\frac{2\pi }{T}$ r = $\frac{2\pi }{0.25}$ × 0.15 = 3.77 ms^–1

Q.12. A particle of mass 0.3 kg vibrates with a period of 2 seconds. If its amplitude is 0.5m, what is its maximum kinetic energy?

Here,

Mass, m = 0.3 kg

Time period, T = 2 sec

Amplitude, r = 0.5 m

Maximum K.E.=?

Method-1:

We have,

K.E._max = $\frac{1}{2}$ m (V_max)² 

Velocity is maximum at mean position, where displacement, y = 0 (minimum)

So, K.E._max = $\frac{1}{2}$ m [$\omega $$\sqrt{{{r}^{2}}-{{0}^{2}}}$]²         Since, v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

Or, K.E._max = $\frac{1}{2}$ m $\omega $² r²         

Or, K.E._max = $\frac{1}{2}$ m×${{\left( \frac{2\pi }{T} \right)}^{2}}$× r²   

Or, K.E._max = $\frac{1}{2}$ × 0.3 ×${{\left( \frac{2\pi }{2} \right)}^{2}}$× (0.5)²   

$\therefore $       K.E._max = 0.37 J

Method-2:

In S.H.M. 

K.E. = $\frac{1}{2}$ m$\omega $² (r² – y²)

For maximum K.E. we have displacement, y = 0 

$\therefore $ K.E._max = $\frac{1}{2}$ m$\omega $² (r² – 0²)

K.E_max. = $\frac{1}{2}$ m $\omega $² r²

K.E._max = $\frac{1}{2}$ m×${{\left( \frac{2\pi }{T} \right)}^{2}}$× r²   

K.E._max = $\frac{1}{2}$ × 0.3 ×${{\left( \frac{2\pi }{2} \right)}^{2}}$× (0.5)²   

$\therefore $     K.E._max = 0.37 J

Method-3:

In S.H.M. 

Maximum K.E. = Total energy 

i.e. K.E._max = T.E. =  m $\omega $² r² 

K.E._max = $\frac{1}{2}$ m×${{\left( \frac{2\pi }{T} \right)}^{2}}$× r²   

K.E._max = $\frac{1}{2}$ × 0.3 ×${{\left( \frac{2\pi }{2} \right)}^{2}}$× (0.5)²   

$\therefore $     K.E._max = 0.37 J

Q.13. A small mass rests on a horizontal platform which vibrates in simple harmonic motion with a period of 0.25s. Find the maximum amplitude of the motion which will allow the mass to remain in contact with the platform throughout the motion. 

Here,

Time period, T = 0.25 sec

Maximum amplitude, r_max =?

For the object to remain in contact with the platform

Maximum acceleration = acceleration due to gravity

i.e. a_max = g

^{$\omega $2} r_max = g

${{\left( \frac{2\pi }{T} \right)}^{2}}$× r_max = g

r_max = ${{\left( \frac{T}{2\pi } \right)}^{2}}$× g = ${{\left( \frac{0.25}{2\pi } \right)}^{2}}$× 10

$\therefore $    r_max = 0.016 m

Q.14. The velocity of the particle executing simple harmonic motion is 16 cms^–1 at a distance of 8 cm from the mean position and 8 cms^–1 at a distance of 12 cm from the mean position. Calculate the amplitude of the motion.

Here,

Case -1:

Velocity, v₁ = 16 cm s^–1 = 0.16 ms^–1

Displacement, y₁ = 8 cm = 0.08 m

Case -2:

Velocity, v₂ = 8 cms^–1  = 0.08 ms^–1  

Displacement, y₁ = 12 cm = 0.08 m

Amplitude, r = ?

We have,

Velocity, v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$

From 1^st case,

0.16 = $\omega $$\sqrt{{{r}^{2}}-{{0.08}^{2}}}$……. (1)

From 2^nd case,

0.08 = $\omega $$\sqrt{{{r}^{2}}-{{0.12}^{2}}}$……. (2)

Dividing equation (1) by (2) we get,

2 = $\sqrt{\frac{{{r}^{2}}{{0.08}^{2}}}{{{r}^{2}}{{0.12}^{2}}}}$

Or, 4r² – 4×0.12² = r² – ×0.08² 

Or, 3r² = – 0.08² + 4×0.12² 

Or, r = 0.1306 m = 13.06 cm

Type – 4: (when displacement equation)

(1) If displacement is, y = r sin$\omega $t

Then 

(i) velocity, v = $\frac{dy}{dt}$

(ii) Acceleration, a = $\frac{dv}{dt}$ = $\frac{d}{dt}$ $\left( \frac{dy}{dt} \right)$= $\frac{{{d}^{2}}y}{d{{t}^{2}}}$ 

In terms of phase angle, displacement, y = r sin ($\omega $t +$\phi $)

Any given displacement equation can be compared with above equation

 and then amplitude (r),  $\omega $, $\phi $, a, v, T, f, K.E, P.E, T.E can be obtained.

Q.15. The displacement y of a mass vibrating with simple harmonic motion is given by y = 20 sin 10$\pi $t. 

Where y is in millimeter and t is in second. What is: (i) amplitude (ii) the period (iii) the velocity at t = 0. 

Here,

Displacement, y = 20 sin 10$\pi $t.  

y is in millimeter and t is in second 

(i) Amplitude, r =?

(ii) Time period, T =?

(iii) Velocity (at t = 0) =?

Comparing above equation with general displacement equation 

y = r sin $\omega $t we get, 

(i) Amplitude, r = 20 mm = 20×10^–3 m

Also, $\omega $ = 10$\pi $

Or, $\frac{2\pi }{T}$ = 10$\pi $

(ii) T = $\frac{2}{10}$ = 0.2 sec

(iii) Velocity, v = $\frac{dy}{dt}$ = $\frac{d}{dt}$ [20 sin(10$\pi $t)]

Or, v = 20 cos(10$\pi $t) ×10$\pi $

At t = 0, velocity, v = 20 cos 0 ×10$\pi $

Or, velocity, v = 628.32 mm s^–1  

$\therefore $         velocity, v = 0.628 m s^–1  

Alternative method:

 In S.H.M. velocity is given by

v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$…….(i)

At t = 0, y = 20 sin 10$\pi $t = 0

From equation (i)

     v = $\omega $$\sqrt{{{r}^{2}}-{{0}^{2}}}$

Or, v = $\frac{2\pi }{T}$ .r

Or, v = $\frac{2\pi }{0.2}$ × 20

Or, v = 628.32 mm/s

$\therefore $  velocity, v = 0.628 m/s