Question

A spring has a force constant k and mass m . The spring hangs vertically and a block of unknown mass is attached to its bottom end: It is known that the mass of the block is much greater than that of the spring. The hanging block stretches the spring the twice its relaxed length. How long (t) would it take for a low amplitude transverse pulse to travel the length of the spring stretched by the hanging block ?

• A. $\sqrt { \frac { 2 \mathrm {~m} } { \mathrm { k } } }$
• B. $\sqrt { \frac { \mathrm { m } } { \mathrm { k } } }$
• C.
• D. $\sqrt { \frac { 2 \mathrm {~m} } { 3 \mathrm { k } } }$

Answer 1

Answer: (A)

$\sqrt { \frac { 2 \mathrm {~m} } { \mathrm { k } } }$

Answer 2

Since mass of spring is small compared to the mass m, the tension force is approximately constant

\ T = mg

kx = mg x = $\frac { \mathrm { mg } } { \mathrm { k } }$

When x-acceleration of the spring

x = 2L given

Where L – Relaxed length.

New length = 2L

= $\frac { 2 \mathrm { mg } } { \mathrm { k } }$

\ mass per unit length

µ = $\frac { \mathrm { mk } } { 2 \mathrm { mg } }$ =

Speed of wave V = $\sqrt { \frac { \mathrm { T } } { \mu } }$

= $\sqrt { \frac { \mathrm { mg } \times 2 \mathrm { mg } } { \mathrm { mk } } }$

V =

time =

= $\frac { 2 \mathrm { mg } } { \mathrm { k } }$ $\frac { \sqrt { \mathrm { k } } } { \sqrt { 2 \mathrm {~m} } }$

time = $\sqrt { \frac { 2 \mathrm {~m} } { \mathrm { k } } }$