A spring having with a spring constant 1200 N m\\(\displaystyle ^{-1}\\) is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Spring constant, \\(k=1200\,m^{-1}\\)Mass, \\(m=3kg\\)Displacement, \\(A=2.0\,cm=0.02m\\)(i) Frequency of oscillation v, is given by the relation:\\(\displaystyle v=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\)where, T is time period\\(\therefore \displaystyle v=\frac{1}{2\times 3.14}\sqrt{\frac{1200}{3}} = 3.18 Hz\\)Hence, the frequency of oscillations is 3.18 cycles per second.(ii) Maximum acceleration (a) is given by the relation:\\(a=\omega^2\,A\\)where,\\(\omega =\\) Angular frequency \\(=\sqrt{\frac{k}{m}}\\)\\(A\\) = maximum displacement\\(\displaystyle \therefore a=\frac{k}{m}A=\frac{1200\times 0.02}{3}=8\, ms^{-2}\\)Hence, the maximum acceleration of the mass is \\(8.0\,m/s^2\\)(iii) Maximum velocity, \\(v_{max}=A\omega\\)\\(\displaystyle =A\sqrt{\frac{k}{m}}=0.02\times \sqrt{\frac{1200}{3}}=0.4\,m/s\\)Hence, the maximum velocity of the mass is 0.4 m/s.

Two masses \\(m_1\\) and \\(m_2\\) are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, \\(m_1\\) is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of \\(m_2\\).

If only mass \\({ m }_{ 2 }\\) is present the extension of spring \\(l\\) is given by \\({ m }_{ 2 }g=kl\Rightarrow l=\cfrac { { m }_{ 2 }g }{ k } \\), where \\(k\\)=Spring constantNow with \\({ m }_{ 1 }\\) and \\({ m }_{ 2 }\\) the extension be \\(l'\\) \\(({ m }_{ 1 }+{ m }_{ 2 })g=kl'\\)\\(l'=\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } \\)When \\({ m }_{ 1 }\\) is removed, \\((l'-1)\\) acts as amplitude of oscillation\\(A=(l'-l)=(\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } )g-\cfrac { { m }_{ 2 }g }{ k } \\)\\(\Rightarrow A=\cfrac { { m }_{ 1 }g }{ k } \\)Now, if \\({ m }_{ 2 }\\) is removed the angular frequency of oscillation \\(\omega =\sqrt { \cfrac { k }{ { m }_{ 2 } } } \\)(This is same as the case of \\({ m }_{ 2 }\\) attach to spring)\\(*\\)N.B: When \\({ m }_{ 1 }\\) is removed the spring mass system will oscillates with the same angular frequency as the case of spring mass system with only mass \\({ m }_{ 2 }\\)

The frequency of vibration \\('f'\\) of a mass \\('m'\\) suspended from a spring of spring constant \\(K\\) is given by a relation of this type, \\(f = Cm^{x}K^{y}\\); where \\(C\\) is a dimensionless quantity. The value of \\(x\\) and \\(y\\) are

The frequency of oscillation spring mass system \\(=\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { k }{ m } } \\)Where, \\(k\\)=spring constantand \\(m\\)=mass with springAnd given frequency, \\(f=c{ m }^{ x }{ k }^{ y }\\)Comparing above two frequency we get \\(x=-\cfrac { 1 }{ 2 } ,y=\cfrac { 1 }{ 2 } \\)

A spring balance has a scale that reads from \\(0\\) to \\(50 kg\\). The length of the scale is \\(20 cm\\). A body suspended from this balance, when displaced and released, oscillates with period of \\(0.6 s\\). What is the weight of the body ?

Maximum mass that the scale can read, \\(M=50kg\\)Maximum displacement of the spring = Length of the scale, \\(l=20cm = 0.2 m\\)Time period, \\(T=0.6s\\)Maximum force exerted on the spring, \\(F=Mg\\)where,\\(g=\\) acceleration due to gravity \\(=9.8\,m/s^2\\)\\(F=50\times 9.8=490\\)\\(\therefore\\) Spring constant, \\(k=F/l=490/0.2=2450\,Nm^{-1}\\)Mass m, is suspended from the balance.Times period, \\(t=2\pi\sqrt{\dfrac{m}{k}}\\)\\(\therefore m=\left(\dfrac{T}{2\pi}\right)^2\times k=\left(\dfrac{0.6}{2\times 3.14}\right)^2\times 2450=22.36kg\\)\\(\therefore\\) Weight of the body \\(=mg=22.36\times 9.8=219.16\,N\\)Hence, the weight of the body is about 219 N.

Two blocks \\(A\\) and \\(B\\) of masses \\(2m\\) and \\(m\\), respectively are connected by a massless and inextensible string. The whole system is syspended by a massless spring as shown in the figure. The magnitude of acceleration of \\(A\\) and \\(B\\) immediately after the string is cut, are respectively:

Two blocks A and B of mass \\(2m\\)and \\(m\\) respectively and they are connected by a mass less and in extensible string.The equilibrium condition FBD of \\(2m\\)\\(Kx=T+2mg\\)\\(=mg+2mg=3mg\\)After the string is cut\\(T=0\\)FBD of \\(2m\\)\\(a_2=\cfrac{3mg-2mg}{2m}\\ \quad=\cfrac{g}{2}\\)FBD of \\(m\\)Hence \\(\cfrac{g}{2}\\) and \\(g\\)

Two identical springs of constant are connected in series and parallel as shown in figure. A mass \\(m\\) is suspended from them. The ratio of their frequencies of vertical oscillations will be

We know that, \\(n=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\)In series, \\(k=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{k^2}{2}\\)In parallel, \\(k=k_1+k_2=2k\\)\\(\therefore, n_1=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}}\\)and \\(n_2 = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}}\\)hence, \\(n_1:n_2=1:2\\)

A spring of force constant \\(K\\) is cut into two pieces such that the one piece is double the length of the other. Then the long piece will have a force constant of:

Force constant, \\( K \propto \dfrac{1}{l}=\dfrac{c}{l}\\)Let the spring cut into pieces of length \\(x, 2x \\)So, \\(x+2x=l \Rightarrow x=\dfrac{l}{3}\\)Force constant of long piece is: \\( k'=\dfrac{c}{2x}=\dfrac{Kl}{2l/3}=\dfrac{3K}{2}\\)

A spring of spring constant 5 x \\(10 ^ { 3 }\\) N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is

Work done is change in potential energyThe spring is initially stretch to 5 cm and again further stretched by 5 cm and further stretched by 5 cm\\(\begin{array}{l} w={ p_{ 10 } }-{ p_{ 5 } } \\ =\frac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\left[ { I{ { \left( { \frac { { 10 } }{ { 100 } } } \right) }^{ 2 } }-{ { \left( { \dfrac { 5 }{ { 100 } } } \right) }^{ 2 } } } \right] \\ =\dfrac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\left( { \dfrac { { 10 } }{ { 100 } } +\dfrac { 5 }{ { 100 } } } \right) \left( { \dfrac { { 10 } }{ { 100 } } -\dfrac { 5 }{ { 100 } } } \right) N-m \\ =\dfrac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\times \dfrac { { 15 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } =2.5\times 15\times 5\times { 10^{ 3 } }\times { 10^{ -4 } } \\ =75\times 2.5\times { 10^{ -1 } }=7.5\times 2.5N-m \\ =18.75\, \, N-m \end{array}\\)Option A.

The time period of mass suspended from a spring is \\(T\\). If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

\\(T = 2\pi \sqrt {\dfrac{m}{k}}\\)When the spring is cut into four equal parts, the spring constant of one part will becomes 4k, therefore the new time period will become\\(T' = 2\pi \sqrt {\dfrac{m}{k'}}\\)\\(T = 2\pi \sqrt {\dfrac{m}{4k}}\\) \\(T'=\dfrac{T}{2}\\)

A mass \\(M\\) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period \\(T\\). If the mass is increased by \\(m\\), the time period becomes \\(\displaystyle\frac{5T}{3}\\). Then the ratio of \\(\displaystyle\frac{m}{M}\\) is

From formula 1, we can say\\(T=2\pi \sqrt{\dfrac{M}{k}}\\) ...(1)Also,\\(\dfrac{5T}{3}=2\pi \sqrt{\dfrac{M+m}{k}}\\) ...(2)Dividing eqn 2 and 1, we get\\(\dfrac{5}{3}= \sqrt{\dfrac{M+m}{M}}\\)\\(\therefore \dfrac{25}{9}= \dfrac{M+m}{M}\\)\\(\therefore \dfrac{25}{9}=1+ \dfrac{m}{M}\\)Hence, answer=16/9, i.e. option B

Problems on SHM of Springs | Definition, Examples, Diagrams

example

Solve problems on describing motion of two blocks attached with a spring

Example: A block of mass

m = 1 k g

placed on top of another block of mass

M = 5 k g

is attached to a horizontal spring of force constant

k = 20 N / m

as shown in figure. The coefficient of friction between the blocks is

μ

where as the lower block slides on a friction less surface. The amplitude oscillation is 0.4 m. What is the minimum value of

μ

such that the upper block does not slip over the lower block?
The upper block does not slip over the lower block when the restoring force is balanced by the friction force of lower block against ground. i.e,

k x_{0} = μ (M + m) g

μ = \frac{k x _{0}}{( M + m ) g} = \frac{2 0 \times 0 . 4}{( 5 + 1 ) 1 0} = 0.133

example

Problem of various combination of Springs

Let the spring constant of the spring be

K

.
Thus the effective spring constant of the parallel combination as shown in the figure=

2 K

Thus in a series combination of above springs, effective spring constant=

K_{i} = \frac{K \times 2 K}{K + 2 K}

= \frac{2 K}{3}

In parallel combination of above springs, effective spring constant=

K_{p} = K + 2 K = 3 K

Thus

\frac{K _{i}}{K _{p}} = \frac{2}{9}

example

Solve problems in which part of the motion is SHM

\frac{W}{k} = x_{1} + x_{2} = \frac{W}{k _{1}} + \frac{W}{k _{2}}

Equivalent spring constant:

k = \frac{k _{1} k _{2}}{k _{1} + k _{2}}

example

SHM in spring in accelerated frame of reference

Example:
A block of mass

1

kg is kept on smooth floor of a truck. One end of a spring of force constant

100

N/m is attached to the block and other end is attached to the body of truck as shown in the figure. At

t = 0

, truck begins to move with constant acceleration

2

m/s

^{2}

. Find the amplitude of oscillation of block relative to the floor of truck.
Solution:
Let

x_{0}

is the compression in equilibrium. Then

k x_{0} = m a

x_{0} = \frac{m a}{k}

= \frac{1 \times 2}{1 0 0}

= 0.02

m
Amplitude

= x_{0}

= 0.02

example

Springs in rotational SHM

Example:
A uniform rod of length l mass m is fixed at the centre. A spring of spring constant k is connected to rod and wall as shown in figure. The rod is displaced by small angle

θ

and released. Find time period of oscillation.
Solution:
Let the length of the rod be

l

.
Torque acting on the rod about the center of the rod is given by

τ = I α = - (k x) \frac{l}{2}

\frac{m l ^{2}}{1 2} α = - k \frac{l}{2} θ (\frac{l}{2})

α = - \frac{3 k}{m} θ

\Rightarrow ω = \frac{3 k}{m}

T = 2 π \frac{m}{3 k}

Problems on SHM of Springs

example

Solve problems on describing motion of two blocks attached with a spring

example

Problem of various combination of Springs

example

Solve problems in which part of the motion is SHM

example

SHM in spring in accelerated frame of reference

example

Springs in rotational SHM

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Oscillations Due to a Spring

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